HC Verma - Wave Motion and Wave on a String Solution For Class 11 Concepts Of Physics Part 1

Question 1:

Answer:

No, in wave motion there is no actual transfer of matter but transfer of energy between the points where as when wind blows air particles moves with it.

Question 2:

No, in wave motion there is no actual transfer of matter but transfer of energy between the points where as when wind blows air particles moves with it.

Answer:

It is a non-mechanical wave because this type of wave does not require a material medium to travel.

Question 3:

It is a non-mechanical wave because this type of wave does not require a material medium to travel.

Answer:

Equation of the wave is
$y = c_{1} \sin (c_{2} x + c_{3} t)$
When the variable of the equation is (c₂x + c₃t), then the wave must be moving in the negative x-axis with time t.

Question 4:

Equation of the wave is
$y = c_{1} \sin (c_{2} x + c_{3} t)$
When the variable of the equation is (c₂x + c₃t), then the wave must be moving in the negative x-axis with time t.

Answer:

Equation of the wave is given by
$y = A \sin (ω t - k x)$
where
  A is the amplitude
           ω is the angular frequency
           k is the wave number
Velocity of wave, $v = \frac{ω}{k}$
Velocity of particle, $v_{p} = \frac{d y}{d t} = A ω \cos (ω t - k x)$
Max velocity of particle, $v_{p_{\max}} = A ω$
As given
$A < \frac{λ}{2 π}$
$v_{p_{\max}} = \frac{λ ω}{2 π} v_{p_{\max}} < \frac{ω}{k} [∵ \frac{2 π}{λ} = k]$

Question 5:

Equation of the wave is given by
$y = A \sin (ω t - k x)$
where
  A is the amplitude
           ω is the angular frequency
           k is the wave number
Velocity of wave, $v = \frac{ω}{k}$
Velocity of particle, $v_{p} = \frac{d y}{d t} = A ω \cos (ω t - k x)$
Max velocity of particle, $v_{p_{\max}} = A ω$
As given
$A < \frac{λ}{2 π}$
$v_{p_{\max}} = \frac{λ ω}{2 π} v_{p_{\max}} < \frac{ω}{k} [∵ \frac{2 π}{λ} = k]$

Answer:

When two wave pulses identical in shape but inverted with respect to each other meet at any instant, they form a destructive interference. The complete energy of the system at that instant is stored in the form of potential energy within it. After passing each other, both the pulses regain their original shape.

Question 6:

When two wave pulses identical in shape but inverted with respect to each other meet at any instant, they form a destructive interference. The complete energy of the system at that instant is stored in the form of potential energy within it. After passing each other, both the pulses regain their original shape.

Answer:

$\frac{y_{\max}}{v_{\max}} = \frac{v_{\max}}{a_{\max}} y = A \sin (ω t - k x) y = A v = \frac{d y}{d t} = A ω \cos (ω t - k x) v_{\max} = A ω a = \frac{d v}{d t} = - A ω^{2} \sin (ω t - k x) a_{\max} = ω^{2} A$

To prove,

$\frac{y_{\max}}{v_{\max}} = \frac{v_{\max}}{a_{\max}} LHS \frac{y_{\max}}{v_{\max}} = \frac{A}{A ω} = \frac{1}{ω} RHS \frac{v_{\max}}{a_{\max}} = \frac{A ω}{ω^{2} A} = \frac{1}{ω}$

No, componendo and dividendo is not applicable. We cannot add quantities of different dimensions.

Question 7:

$\frac{y_{\max}}{v_{\max}} = \frac{v_{\max}}{a_{\max}} y = A \sin (ω t - k x) y = A v = \frac{d y}{d t} = A ω \cos (ω t - k x) v_{\max} = A ω a = \frac{d v}{d t} = - A ω^{2} \sin (ω t - k x) a_{\max} = ω^{2} A$

To prove,

$\frac{y_{\max}}{v_{\max}} = \frac{v_{\max}}{a_{\max}} LHS \frac{y_{\max}}{v_{\max}} = \frac{A}{A ω} = \frac{1}{ω} RHS \frac{v_{\max}}{a_{\max}} = \frac{A ω}{ω^{2} A} = \frac{1}{ω}$

No, componendo and dividendo is not applicable. We cannot add quantities of different dimensions.

Answer:

Equation of the wave: y = A sin(kx − ωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.

The argument of the sine is a phase, so the smallest positive phase constant should be
$\sin (7.5 π) = \sin (3 \times 2 π + 1.5 π) = \sin (1.5 π)$
Therefore, the smallest positive phase constant is 1.5π.

Question 8:

Equation of the wave: y = A sin(kx − ωt + Φ)
Here, A is the amplitude, k is the wave number, ω is the angular frequency and Φ is the initial phase.

The argument of the sine is a phase, so the smallest positive phase constant should be
$\sin (7.5 π) = \sin (3 \times 2 π + 1.5 π) = \sin (1.5 π)$
Therefore, the smallest positive phase constant is 1.5π.

Answer:

Yes, at the centre. The centre position is a node. If the string vibrates in its first overtone, then there will be two positions, i.e., two nodes, one at x = 0 and the other at x = L.

Question 1:

Yes, at the centre. The centre position is a node. If the string vibrates in its first overtone, then there will be two positions, i.e., two nodes, one at x = 0 and the other at x = L.

Answer:

(c) $λ / 2$

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the positions of maxima and minima is equal to $λ / 2$ .

Question 2:

(c) $λ / 2$

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. The speeds at the maximum point and at the minimum point are same although the direction of motion are different. The difference between the positions of maxima and minima is equal to $λ / 2$ .

Answer:

(c) $λ / 2$

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant, then the particle closest to it having zero displacement is at a distance is equal to $λ / 2$ .

Question 3:

(c) $λ / 2$

A sine wave has a maxima and a minima and the particle displacement has phase difference of π radians. Therefore, applying similar argument we can say that if a particular particle has zero displacement at a certain instant, then the particle closest to it having zero displacement is at a distance is equal to $λ / 2$ .

Answer:

(a) $x = A \sin (k y - ω t)$

Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.

Question 4:

(a) $x = A \sin (k y - ω t)$

Here x is the particle displacement of the wave and the wave is travelling along the Y-axis because the particle displacement is perpendicular to the direction of wave motion.

Answer:

(b) amplitude A/2, frequency $ω / π$

$y = A \sin^{2} (k x - ω t)$

$[\cos^{2} θ = 1 - 2 \sin^{2} θ \sin^{2} θ = \frac{1 - \cos^{2} θ}{2}]$

$y = A [\frac{1 - \cos^{2} (k x - ω t)}{2}] y = \frac{A}{2} [1 - \cos^{2} (k x - ω t)]$
Thus, we have:
Amplitude = $\frac{A}{2}$
Frequency = $2 (\frac{ω}{2 π}) = \frac{ω}{π}$

Question 5:

(b) amplitude A/2, frequency $ω / π$

$y = A \sin^{2} (k x - ω t)$

$[\cos^{2} θ = 1 - 2 \sin^{2} θ \sin^{2} θ = \frac{1 - \cos^{2} θ}{2}]$

$y = A [\frac{1 - \cos^{2} (k x - ω t)}{2}] y = \frac{A}{2} [1 - \cos^{2} (k x - ω t)]$
Thus, we have:
Amplitude = $\frac{A}{2}$
Frequency = $2 (\frac{ω}{2 π}) = \frac{ω}{π}$

Answer:

(d) Sound waves

There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound requires medium to travel, hence it is a mechanical wave.

Question 6:

(d) Sound waves

There are mainly two types of waves: first is electromagnetic wave, which does not require any medium to travel, and the second is the mechanical wave, which requires a medium to travel. Sound requires medium to travel, hence it is a mechanical wave.

Answer:

(a) $ν$

The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.

Question 7:

(a) $ν$

The boat transmits the same wave without any change of frequency to cause the cork to execute SHM with same frequency though amplitude may differ.

Answer:

(a) 1/2

Wave speed is given by
$ν = \sqrt{\frac{T}{µ}}$
where
             â€‹T is the tension in the string
             v is the speed of the wave
             μ is the mass per unit length of the string
$µ = \frac{M}{L} = ρ \frac{V}{L} = ρ \frac{(A L)}{L}$
where
M is the mass of the string, which can be written as ρV
             L is the length of the string
$= ρ (π r^{2}) = ρ (π \frac{D^{2}}{4}) ∴ ν = \sqrt{\frac{T}{ρ π \frac{D^{2}}{4}}} = \frac{2}{D} \sqrt{\frac{T}{ρ π}}$
where D is the diameter of the string.
Thus, v ∝ $\frac{1}{D}$
Since, r_A = 2r_B
$v_{A} \propto \frac{1}{2 r_{A}} \propto \frac{1}{2 \times 2 r_{B}} (1) v_{_{B}} \propto \frac{1}{2 r_{_{B}}} (2)$
From Equations (1) and (2) we get
$\frac{v_{A}}{v_{B}} = \frac{1}{2}$

Question 8:

(a) 1/2

Wave speed is given by
$ν = \sqrt{\frac{T}{µ}}$
where
             â€‹T is the tension in the string
             v is the speed of the wave
             μ is the mass per unit length of the string
$µ = \frac{M}{L} = ρ \frac{V}{L} = ρ \frac{(A L)}{L}$
where
M is the mass of the string, which can be written as ρV
             L is the length of the string
$= ρ (π r^{2}) = ρ (π \frac{D^{2}}{4}) ∴ ν = \sqrt{\frac{T}{ρ π \frac{D^{2}}{4}}} = \frac{2}{D} \sqrt{\frac{T}{ρ π}}$
where D is the diameter of the string.
Thus, v ∝ $\frac{1}{D}$
Since, r_A = 2r_B
$v_{A} \propto \frac{1}{2 r_{A}} \propto \frac{1}{2 \times 2 r_{B}} (1) v_{_{B}} \propto \frac{1}{2 r_{_{B}}} (2)$
From Equations (1) and (2) we get
$\frac{v_{A}}{v_{B}} = \frac{1}{2}$

Answer:

(d) $1 / \sqrt{2}$

$T_{AB} = T T_{CD} = 2 T$
where
          â€‹T_AB is the tension in the string AB
          â€‹T_CD is the tension in the string CD
The eelation between tension and the wave speed is given by
$v = \sqrt{\frac{T}{µ}} v \propto \sqrt{T}$
where
  v is the wave speed of the transverse wave
           μ is the mass per unit length of the string
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T}{2 T}} = \frac{1}{\sqrt{2}}$

Question 9:

(d) $1 / \sqrt{2}$

$T_{AB} = T T_{CD} = 2 T$
where
          â€‹T_AB is the tension in the string AB
          â€‹T_CD is the tension in the string CD
The eelation between tension and the wave speed is given by
$v = \sqrt{\frac{T}{µ}} v \propto \sqrt{T}$
where
  v is the wave speed of the transverse wave
           μ is the mass per unit length of the string
$\frac{v_{1}}{v_{2}} = \sqrt{\frac{T}{2 T}} = \frac{1}{\sqrt{2}}$

Answer:

(d) meaningless

Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.

Question 10:

(d) meaningless

Sound wave is a mechanical wave; this means that it needs a medium to travel. Thus, its velocity in vacuum is meaningless.

Answer:

(c) $λ' < λ$

As $v = \sqrt{\frac{f}{µ}}$
A wave pulse travels faster in a thinner string.
The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.

Question 11:

(c) $λ' < λ$

As $v = \sqrt{\frac{f}{µ}}$
A wave pulse travels faster in a thinner string.
The wavelength of the transmitted wave is equal to the wavelength of the incident wave because the frequency remains constant.

Answer:

(b) $\sqrt{2} a$

We know that the resultant of the amplitude is given by
$R_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
For the particular case, we can write
$= \sqrt{a^{2} + a^{2} + 2 a^{2} \cos \frac{π}{2}} = \sqrt{2} a$

Question 12:

(b) $\sqrt{2} a$

We know that the resultant of the amplitude is given by
$R_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
For the particular case, we can write
$= \sqrt{a^{2} + a^{2} + 2 a^{2} \cos \frac{π}{2}} = \sqrt{2} a$

Answer:

(d) the information is insufficient to find the relation between t₁ and t₂.

$v = \sqrt{\frac{η}{ρ}}$
But because the length of wires A and B is not known, the relation between A and B cannot be determined.

Question 13:

(d) the information is insufficient to find the relation between t₁ and t₂.

$v = \sqrt{\frac{η}{ρ}}$
But because the length of wires A and B is not known, the relation between A and B cannot be determined.

Answer:

(b) the velocity but not for the kinetic energy

The principle of superposition is valid only for vector quantities. Velocity is a vector quantity, but kinetic energy is a scalar quantity.

Question 14:

(b) the velocity but not for the kinetic energy

The principle of superposition is valid only for vector quantities. Velocity is a vector quantity, but kinetic energy is a scalar quantity.

Answer:

(d) The pulses will pass through each other without any change in their shapes.

The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.

Question 15:

(d) The pulses will pass through each other without any change in their shapes.

The pulses continue to retain their identity after they meet, but the moment they meet their wave profile differs from the individual pulse.

Answer:

(b) 2A₂

We know resultant amplitude is given by
$A_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
For maximum resultant amplitude
$A_{\max} = A_{1} + A_{2}$
For minimum resultant amplitude
$A_{\min} = A_{1} - A_{2}$

So, the difference between A_max and A_min is
$A_{\max} - A_{\min} = A_{1} + A_{2} - A_{1} + A_{2} = 2 A_{2}$

Question 16:

(b) 2A₂

We know resultant amplitude is given by
$A_{net} = \sqrt{A_{1}^{2} + A_{2}^{2} + 2 A_{1} A_{2} \cos ϕ}$
For maximum resultant amplitude
$A_{\max} = A_{1} + A_{2}$
For minimum resultant amplitude
$A_{\min} = A_{1} - A_{2}$

So, the difference between A_max and A_min is
$A_{\max} - A_{\min} = A_{1} + A_{2} - A_{1} + A_{2} = 2 A_{2}$

Answer:

(d) between 0 and 2A

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle (φ). So, the resultant amplitude lies between the maximum resultant amplitude (A_max) and the minimum resultant amplitude (A_min).
A_max = A + A = 2A
A_min= A − A = 0

Question 17:

(d) between 0 and 2A

The amplitude of the resultant wave depends on the way two waves superimpose, i.e., the phase angle (φ). So, the resultant amplitude lies between the maximum resultant amplitude (A_max) and the minimum resultant amplitude (A_min).
A_max = A + A = 2A
A_min= A − A = 0

Answer:

(a) A

We know the resultant amplitude is given by
$R_{net} = \sqrt{A^{2} + A^{2} + 2 A^{2} \cos 120 º} (ϕ = 120 °) = \sqrt{2 A^{2} - A^{2}} [∵ \cos 120 º = \frac{- 1}{2}] = A$

Question 18:

(a) A

We know the resultant amplitude is given by
$R_{net} = \sqrt{A^{2} + A^{2} + 2 A^{2} \cos 120 º} (ϕ = 120 °) = \sqrt{2 A^{2} - A^{2}} [∵ \cos 120 º = \frac{- 1}{2}] = A$

Answer:

(a) inverse of its length

The relation between wave speed and the length of the string is given by
$v = \frac{1}{2 l} \sqrt{\frac{F}{µ}}$
where
l is the length of the string
           F is the tension
           μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
$v \propto \frac{1}{l}$

Question 1:

(a) inverse of its length

The relation between wave speed and the length of the string is given by
$v = \frac{1}{2 l} \sqrt{\frac{F}{µ}}$
where
l is the length of the string
           F is the tension
           μ linear mass density
From the above relation, we can say that the fundamental frequency of a string is proportional to the inverse of the length of the string.
$v \propto \frac{1}{l}$

Answer:

Given,
Speed of the wave pulse passing on a string in the negative x-direction = 40 cms⁻¹
As the speed of the wave is constant, the location of the maximum after 5 s will be
s = v × t
= 40 × 5
= 200 cm (along the negative x-axis)
Therefore, the required maximum will be located after x = −2 m.

Question 2:

Given,
Speed of the wave pulse passing on a string in the negative x-direction = 40 cms⁻¹
As the speed of the wave is constant, the location of the maximum after 5 s will be
s = v × t
= 40 × 5
= 200 cm (along the negative x-axis)
Therefore, the required maximum will be located after x = −2 m.

Answer:

Given,
Equation of the wave travelling on a string stretched along the X-axis:
$y = A e^{(\frac{x}{a} + \frac{t}{T}) {^{-}}^{2}}$
(a) The dimensions of A (amplitude), T (time period) and $a = \frac{λ}{2 π}$ , which will have the dimensions of the wavelength, are as follows:
$[A] = [M^{0} L^{1} T^{0}] [T] = [M^{0} L^{0} T^{- 1}] [a] = [M^{0} L^{1} T^{0}]$

(b) Wave speed, $ν = \frac{λ}{T} = \frac{a}{T} [λ = a]$

(c) If $y = f (t + \frac{x}{ν})$ , then the wave travels in the negative direction; and if $y = f (t - \frac{x}{ν})$ , then the wave travels in the positive direction.
Thus, we have:
$y = A e^{{[(\frac{x}{a}) + (\frac{t}{T})]}^{-}^{2}} = A e^{- \frac{1}{T} {[t + (\frac{x T}{a})]}^{2}} = A e^{- \frac{1}{T} [t + \frac{x}{V}]} = A e^{- f [t + \frac{x}{V}]}$
Hence, the wave is travelling is the negative direction.

(d) Wave speed, $v = \frac{a}{t}$
Maximum pulse at t = T = $(\frac{a}{T}) \times T = a (Along the negative x - axis)$
Maximum pulse at t = 2T = $(\frac{a}{T} \times 2 T) = 2 a (Along the negative x - axis)$
Therefore, the wave is travelling in the negative x-direction.

Question 3:

Given,
Equation of the wave travelling on a string stretched along the X-axis:
$y = A e^{(\frac{x}{a} + \frac{t}{T}) {^{-}}^{2}}$
(a) The dimensions of A (amplitude), T (time period) and $a = \frac{λ}{2 π}$ , which will have the dimensions of the wavelength, are as follows:
$[A] = [M^{0} L^{1} T^{0}] [T] = [M^{0} L^{0} T^{- 1}] [a] = [M^{0} L^{1} T^{0}]$

(b) Wave speed, $ν = \frac{λ}{T} = \frac{a}{T} [λ = a]$

(c) If $y = f (t + \frac{x}{ν})$ , then the wave travels in the negative direction; and if $y = f (t - \frac{x}{ν})$ , then the wave travels in the positive direction.
Thus, we have:
$y = A e^{{[(\frac{x}{a}) + (\frac{t}{T})]}^{-}^{2}} = A e^{- \frac{1}{T} {[t + (\frac{x T}{a})]}^{2}} = A e^{- \frac{1}{T} [t + \frac{x}{V}]} = A e^{- f [t + \frac{x}{V}]}$
Hence, the wave is travelling is the negative direction.

(d) Wave speed, $v = \frac{a}{t}$
Maximum pulse at t = T = $(\frac{a}{T}) \times T = a (Along the negative x - axis)$
Maximum pulse at t = 2T = $(\frac{a}{T} \times 2 T) = 2 a (Along the negative x - axis)$
Therefore, the wave is travelling in the negative x-direction.

Answer:

Given,
Wave pulse at t = 0

Wave speed = 10 cms⁻¹
Using the formula $s = v \times t$ , we get:

$At : t = 1 s, s_{1} = ν \times t = 10 \times 1 = 10 cm t = 2 s, s_{2} = ν \times t = 10 \times 2 = 20 cm t = 3 s, s_{3} = ν \times t = 10 \times 3 = 30 cm$

Question 19:

Given,
Wave pulse at t = 0

Wave speed = 10 cms⁻¹
Using the formula $s = v \times t$ , we get:

$At : t = 1 s, s_{1} = ν \times t = 10 \times 1 = 10 cm t = 2 s, s_{2} = ν \times t = 10 \times 2 = 20 cm t = 3 s, s_{3} = ν \times t = 10 \times 3 = 30 cm$

Answer:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.

Question 20:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be natural frequency of the wire, then standing waves with large amplitude are set up in it.

Answer:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.

Question 21:

(b) 480 Hz

The frequency of vibration of a sonometer wire is the same as that of a fork. If this happens to be the natural frequency of the wire, standing waves with large amplitude are set in it.

Answer:

(b) vibrate with a frequency of 208 Hz

According to the relation of the fundamental frequency of a string
$ν = \frac{1}{2 l} \sqrt{\frac{F}{μ}}$
where
  l is the length of the string
           F is the tension
           μ is the linear mass density

We know that ν₁ = 416 Hz, l₁ = l and l₂ = 2l.
$v_{1} \propto \frac{1}{l_{1}} v_{1} l_{1} = v_{2} l_{2} (416) l = v_{2} (2 l) v_{2} = 208 Hz$

Question 22:

(b) vibrate with a frequency of 208 Hz

According to the relation of the fundamental frequency of a string
$ν = \frac{1}{2 l} \sqrt{\frac{F}{μ}}$
where
  l is the length of the string
           F is the tension
           μ is the linear mass density

We know that ν₁ = 416 Hz, l₁ = l and l₂ = 2l.
$v_{1} \propto \frac{1}{l_{1}} v_{1} l_{1} = v_{2} l_{2} (416) l = v_{2} (2 l) v_{2} = 208 Hz$

Answer:

(d) 16 kg

According to the relation of the fundamental frequency of a string
$ν = \frac{1}{2 l} \sqrt{\frac{F}{μ}}$
where l is the length of the string
F is the tension
μ is the linear mass density of the string
We know that ν₁ = 416 Hz, l₁ = l and l₂ = 2l.
Also, m₁ = 4 kg and m₂ = ?
$ν_{1} = \frac{1}{2 l_{1}} \sqrt{\frac{m_{1} g}{μ}} (1)$
$ν_{2} = \frac{1}{2 l_{2}} \sqrt{\frac{m_{2} g}{μ}} (2)$
So, in order to maintain the same fundamental mode
$ν_{1} = ν_{2}$
squaring both sides of equations (1) and (2) and then equating
$\frac{1}{4 l^{2}} \frac{4 g}{μ} = \frac{1}{16 l^{2}} \frac{m_{2} g}{μ} \Rightarrow m_{2} = 16 kg$

Question 1:

(d) 16 kg

According to the relation of the fundamental frequency of a string
$ν = \frac{1}{2 l} \sqrt{\frac{F}{μ}}$
where l is the length of the string
F is the tension
μ is the linear mass density of the string
We know that ν₁ = 416 Hz, l₁ = l and l₂ = 2l.
Also, m₁ = 4 kg and m₂ = ?
$ν_{1} = \frac{1}{2 l_{1}} \sqrt{\frac{m_{1} g}{μ}} (1)$
$ν_{2} = \frac{1}{2 l_{2}} \sqrt{\frac{m_{2} g}{μ}} (2)$
So, in order to maintain the same fundamental mode
$ν_{1} = ν_{2}$
squaring both sides of equations (1) and (2) and then equating
$\frac{1}{4 l^{2}} \frac{4 g}{μ} = \frac{1}{16 l^{2}} \frac{m_{2} g}{μ} \Rightarrow m_{2} = 16 kg$

Answer:

(c) may move on the X-axis
(d) may move on the Y-axis

A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.

Question 2:

(c) may move on the X-axis
(d) may move on the Y-axis

A mechanical wave is of two types: longitudinal and transverse. So, a particle of a mechanical wave may move perpendicular or along the direction of motion of the wave.

Answer:

(d) in the X–Y plane

In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the X–Y plane.

Question 3:

(d) in the X–Y plane

In a transverse wave, particles move perpendicular to the direction of motion of the wave. In other words, if a wave moves along the Z-axis, the particles will move in the X–Y plane.

Answer:

(d) be polarised

A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.

Question 4:

(d) be polarised

A longitudinal wave has particle displacement along its direction of motion; thus, it cannot be polarised.

Answer:

(b) may be longitudinal
(d) may be transverse

Particles in a solid are very close to each other; thus, both longitudinal and transverse waves can travel through it.

Question 5:

(b) may be longitudinal
(d) may be transverse

Particles in a solid are very close to each other; thus, both longitudinal and transverse waves can travel through it.

Answer:

(a) must be longitudinal

Because particles in a gas are far apart, only longitudinal wave can travel through it.

Question 6:

(a) must be longitudinal

Because particles in a gas are far apart, only longitudinal wave can travel through it.

Answer:

(b) A and B move in opposite directions.
(d) The displacements at A and B have equal magnitudes.

A and B have a phase difference of π. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.
$\vec{y_{A}} = A \sin (ω t)$
$\vec{y_{B}} = B \sin (ω t + π)$

Question 7:

(b) A and B move in opposite directions.
(d) The displacements at A and B have equal magnitudes.

A and B have a phase difference of π. So, when a sine wave passes through the region, they move in opposite directions and have equal displacement. They may be separated by any odd multiple of their wavelength.
$\vec{y_{A}} = A \sin (ω t)$
$\vec{y_{B}} = B \sin (ω t + π)$

Answer:

(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm

$y = (0 \cdot 001 mm) \sin [(50 s^{- 1}) t + (2 \cdot 0 m^{- 1}) x]$
Equating the above equation with the general equation, we get:
$y = A \sin (ω t - k x) ω = \frac{2 π}{T} = 2 π v k = \frac{2 π}{λ}$
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.
$A = 0.001 mm Now, 50 = 2 π ν \Rightarrow ν = \frac{25}{π} Hz$

Question 8:

(c) Frequency = 25/π Hz
(d) Amplitude = 0⋅001 mm

$y = (0 \cdot 001 mm) \sin [(50 s^{- 1}) t + (2 \cdot 0 m^{- 1}) x]$
Equating the above equation with the general equation, we get:
$y = A \sin (ω t - k x) ω = \frac{2 π}{T} = 2 π v k = \frac{2 π}{λ}$
Here, A is the amplitude, ω is the angular frequency, k is the wave number and λ is the wavelength.
$A = 0.001 mm Now, 50 = 2 π ν \Rightarrow ν = \frac{25}{π} Hz$

Answer:

(a) must be an integral multiple of $λ / 4$

A standing wave is produced on a string clamped at one end and free at the other. Its fundamental frequency is given by
$ν = (n + \frac{1}{2}) \frac{v}{2 L} \Rightarrow v = ν λ \Rightarrow ν = (n + \frac{1}{2}) \frac{ν λ}{2 L} \Rightarrow L = (\frac{2 n + 1}{4}) λ \Rightarrow L = \frac{λ}{4}, \frac{3 λ}{4}, . . .$

Question 9:

(a) must be an integral multiple of $λ / 4$

A standing wave is produced on a string clamped at one end and free at the other. Its fundamental frequency is given by
$ν = (n + \frac{1}{2}) \frac{v}{2 L} \Rightarrow v = ν λ \Rightarrow ν = (n + \frac{1}{2}) \frac{ν λ}{2 L} \Rightarrow L = (\frac{2 n + 1}{4}) λ \Rightarrow L = \frac{λ}{4}, \frac{3 λ}{4}, . . .$

Answer:

(b) The energy of any small part of a string remains constant in a standing wave.

A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.

Question 10:

(b) The energy of any small part of a string remains constant in a standing wave.

A standing wave is formed when the energy of any small part of a string remains constant. If it does not, then there is transfer of energy. In that case, the wave is not stationary.

Answer:

(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase

All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.

Thus, particles in alternate antinodes vibrate in the same phase.

Question 4:

(c) the alternate antinodes vibrate in phase
(d) all the particles between consecutive nodes vibrate in phase

All particles in a particular segment between two nodes vibrate in the same phase, but the particles in the neighbouring segments vibrate in opposite phases, as shown below.

Thus, particles in alternate antinodes vibrate in the same phase.

Answer:

Given,
Pulse travelling on a string, $y = [\frac{{(a)}^{3}}{{(x - ν t)}^{2} + a^{2}}]$
$a = 5 mm = 0.5 cm Wave speed, ν = 20 cm / s$
So, at $t = 0 s, y = \frac{a^{3}}{(x^{2} + a^{2})}$ .
Similarly, at t = 1 s,
$y = \frac{a^{3}}{{(x - ν)}^{2} + a^{2}} And, At t = 2 s, y = \frac{a^{3}}{{(x - 2 ν)}^{2} + a^{2}}$

To sketch the shape of the string, we have to plot a graph between y and x at different values of t.

Question 5:

Given,
Pulse travelling on a string, $y = [\frac{{(a)}^{3}}{{(x - ν t)}^{2} + a^{2}}]$
$a = 5 mm = 0.5 cm Wave speed, ν = 20 cm / s$
So, at $t = 0 s, y = \frac{a^{3}}{(x^{2} + a^{2})}$ .
Similarly, at t = 1 s,
$y = \frac{a^{3}}{{(x - ν)}^{2} + a^{2}} And, At t = 2 s, y = \frac{a^{3}}{{(x - 2 ν)}^{2} + a^{2}}$

To sketch the shape of the string, we have to plot a graph between y and x at different values of t.

Answer:

Given,
Equation of the wave travelling in the positive x-direction at x = 0:
$f (t) = A \sin (\frac{t}{T})$
Here,
Wave speed = v
Wavelength, λ = vT
T = Time period
Therefore, the general equation of the wave can be represented by
$y = A \sin [(\frac{t}{T}) - (\frac{x}{ν T})]$

Question 6:

Given,
Equation of the wave travelling in the positive x-direction at x = 0:
$f (t) = A \sin (\frac{t}{T})$
Here,
Wave speed = v
Wavelength, λ = vT
T = Time period
Therefore, the general equation of the wave can be represented by
$y = A \sin [(\frac{t}{T}) - (\frac{x}{ν T})]$

Answer:

The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
Dimensions of A and a are governed by the dimensional homogeneity of the equation g(x) = A sin(x/a).
Now,
$(a) [M^{0} L^{1} T^{0}] = [A] \Rightarrow [A] = [L] And, [a] = [M^{0} L^{1} T^{0}] \Rightarrow [a] = [L] (b) Wave speed = ν ∴ Time period, T = \frac{a}{ν} Here, a = Wave length = λ The general equation of wave is represented by y = A \sin \{\frac{x}{a} - \frac{t}{\frac{a}{v}}\} = A \sin \{\frac{x - ν t}{a}\}$

Question 7:

The shape of the string at t = 0 is given by g(x) = A sin(x/a), where A and a are constants.
Dimensions of A and a are governed by the dimensional homogeneity of the equation g(x) = A sin(x/a).
Now,
$(a) [M^{0} L^{1} T^{0}] = [A] \Rightarrow [A] = [L] And, [a] = [M^{0} L^{1} T^{0}] \Rightarrow [a] = [L] (b) Wave speed = ν ∴ Time period, T = \frac{a}{ν} Here, a = Wave length = λ The general equation of wave is represented by y = A \sin \{\frac{x}{a} - \frac{t}{\frac{a}{v}}\} = A \sin \{\frac{x - ν t}{a}\}$

Answer:

Given,
Wave velocity = $ν$
Shape of the string at $t = t_{0}$ = $g (x, t_{0}) = A \sin (x / a)$ ...(i)
For a wave travelling in the positive x-direction, the general equation is given by
$y = A \sin (\frac{x}{a} - \frac{t}{T})$
Putting t = − t and comparing with equation (i), we get:
$g (x, 0) = A \sin \{(\frac{x}{a}) + (\frac{t_{0}}{T})\} \Rightarrow g (x, t) = A \sin [\{(\frac{x}{a}) + \frac{t_{0}}{T}\} - (\frac{t}{T})] Now, T = \frac{a}{ν} Here, a = Wave length ν = Velocity of the wave Thus, we have : y = A \sin [(\frac{x}{a}) + \frac{t_{0}}{(\frac{a}{ν})} - \frac{t}{(\frac{a}{ν})}]$

$\Rightarrow y = A \sin \frac{x + ν (t_{0} - t)}{a}$

Question 8:

Given,
Wave velocity = $ν$
Shape of the string at $t = t_{0}$ = $g (x, t_{0}) = A \sin (x / a)$ ...(i)
For a wave travelling in the positive x-direction, the general equation is given by
$y = A \sin (\frac{x}{a} - \frac{t}{T})$
Putting t = − t and comparing with equation (i), we get:
$g (x, 0) = A \sin \{(\frac{x}{a}) + (\frac{t_{0}}{T})\} \Rightarrow g (x, t) = A \sin [\{(\frac{x}{a}) + \frac{t_{0}}{T}\} - (\frac{t}{T})] Now, T = \frac{a}{ν} Here, a = Wave length ν = Velocity of the wave Thus, we have : y = A \sin [(\frac{x}{a}) + \frac{t_{0}}{(\frac{a}{ν})} - \frac{t}{(\frac{a}{ν})}]$

$\Rightarrow y = A \sin \frac{x + ν (t_{0} - t)}{a}$

Answer:

Given,
Equation of the wave, $y = (0.10 mm) \sin (31.4 m^{- 1}) x + (314 s^{- 1}) t$
The general equation is $y = A \sin \{(\frac{2 π x}{λ}) + ω t\}$ .
From the above equation, we can conclude:

(a) The wave is travelling in the negative x-direction.

(b) $\frac{2 π}{λ} = 31.4 m^{- 1}$
$\Rightarrow λ = \frac{2 π}{31.4} = 0.2 m = 20 cm And, ω = 314 s^{- 1} \Rightarrow 2 π f = 314 \Rightarrow f = \frac{314}{2 π} = \frac{314}{2 \times 3.14} = 50 s^{- 1} = 50 Hz$

Wave speed:
$ν = λ f = 20 \times 50 = 1000 cm / s$

(c) Maximum displacement, A = 0.10 mm
$Maximum velocity = a ω = 0.1 \times 10^{- 1} \times 314 = 3.14 cm / s$

Question 9:

Given,
Equation of the wave, $y = (0.10 mm) \sin (31.4 m^{- 1}) x + (314 s^{- 1}) t$
The general equation is $y = A \sin \{(\frac{2 π x}{λ}) + ω t\}$ .
From the above equation, we can conclude:

(a) The wave is travelling in the negative x-direction.

(b) $\frac{2 π}{λ} = 31.4 m^{- 1}$
$\Rightarrow λ = \frac{2 π}{31.4} = 0.2 m = 20 cm And, ω = 314 s^{- 1} \Rightarrow 2 π f = 314 \Rightarrow f = \frac{314}{2 π} = \frac{314}{2 \times 3.14} = 50 s^{- 1} = 50 Hz$

Wave speed:
$ν = λ f = 20 \times 50 = 1000 cm / s$

(c) Maximum displacement, A = 0.10 mm
$Maximum velocity = a ω = 0.1 \times 10^{- 1} \times 314 = 3.14 cm / s$

Answer:

A wave travels along the positive x-direction.
Wave amplitude (A) = 0.20 cm
Wavelength (λ) = 20 cm
Wave speed (v) = 20 m/s

(a) General wave equation along the x-axis:
$y = A \sin (k x - ω t) ∴ k = \frac{2 π}{λ} = \frac{2 π}{2} = π {cm}^{- 1} T = \frac{λ}{ν} = \frac{2}{2000} = \frac{1}{1000} = 10^{- 3} s ω = \frac{2 π}{T} = 2 π \times 10^{3} s^{- 1}$
Wave equation:
$y = (0.2 cm) \sin [(π {cm}^{- 1}) x - (2 π \times 10^{- 3} s^{- 1})]$

(b) As per the question
For the wave equation ,we need to find the displacement and velocity at x = 2 cm and t = 0.
$y = (0.2) cm \sin 2 π = 0 ∴ ν = A ω cosπ x = 0.2 \times 2000 π \times \cos 2 π = 400 π = 400 π cm / s = 4 π m / s$

If the wave equation is written in a different fashion, then also we will get the same values for these quantities.

Question 10:

A wave travels along the positive x-direction.
Wave amplitude (A) = 0.20 cm
Wavelength (λ) = 20 cm
Wave speed (v) = 20 m/s

(a) General wave equation along the x-axis:
$y = A \sin (k x - ω t) ∴ k = \frac{2 π}{λ} = \frac{2 π}{2} = π {cm}^{- 1} T = \frac{λ}{ν} = \frac{2}{2000} = \frac{1}{1000} = 10^{- 3} s ω = \frac{2 π}{T} = 2 π \times 10^{3} s^{- 1}$
Wave equation:
$y = (0.2 cm) \sin [(π {cm}^{- 1}) x - (2 π \times 10^{- 3} s^{- 1})]$

(b) As per the question
For the wave equation ,we need to find the displacement and velocity at x = 2 cm and t = 0.
$y = (0.2) cm \sin 2 π = 0 ∴ ν = A ω cosπ x = 0.2 \times 2000 π \times \cos 2 π = 400 π = 400 π cm / s = 4 π m / s$

If the wave equation is written in a different fashion, then also we will get the same values for these quantities.

Answer:

The wave equation is represented by
$y = (1 \cdot 0 mm) \sin π (\frac{x}{2 \cdot 0 cm} - \frac{t}{0 \cdot 01 s})$
Let:
Time period = T
Wavelength = λ

$(a) T = 2 \times 0.01 = 0.02 s = 20 ms λ = 2 \times 2 = 4 cm$

(b) Equation for the velocity of the particle:
$ν = \frac{d y}{d t} = \frac{d}{d t} [\sin 2 π (\frac{x}{4} - \frac{t}{0.02})]$
$= - 0.50 \cos 2 π \{(\frac{x}{4}) - (\frac{t}{0.02})\} \times \frac{1}{0.02} \Rightarrow ν = - 0.50 \cos 2 π \{(\frac{x}{4}) - (\frac{t}{0.02})\} At x = 1 and t = 0.01 s, ν = - 0.50 \cos 2 π \{\frac{1}{4} - \frac{1}{2}\} = 0 .$

(c) (i) Speed of the particle:
$At x = 3 cm and t = 0.01 s,$ $ν = - 0.50 \cos 2 π \{\frac{3}{4} - \frac{1}{2}\} = 0$ .
(ii) $At x = 5 cm and t = 0.01 s,$
$ν = 0 (iii) At x = 7 cm and t = 0.1 s, ν = 0 . (iv) At x = 1 cm and t = 0.011 s, ν = 50 \cos 2 π \{(\frac{1}{4}) - (\frac{0.011}{0.02})\} = - 50 \cos (\frac{3 π}{5}) = - 9.7 cm / s$
(By changing the value of t, the other two can be calculated.)

Question 11:

The wave equation is represented by
$y = (1 \cdot 0 mm) \sin π (\frac{x}{2 \cdot 0 cm} - \frac{t}{0 \cdot 01 s})$
Let:
Time period = T
Wavelength = λ

$(a) T = 2 \times 0.01 = 0.02 s = 20 ms λ = 2 \times 2 = 4 cm$

(b) Equation for the velocity of the particle:
$ν = \frac{d y}{d t} = \frac{d}{d t} [\sin 2 π (\frac{x}{4} - \frac{t}{0.02})]$
$= - 0.50 \cos 2 π \{(\frac{x}{4}) - (\frac{t}{0.02})\} \times \frac{1}{0.02} \Rightarrow ν = - 0.50 \cos 2 π \{(\frac{x}{4}) - (\frac{t}{0.02})\} At x = 1 and t = 0.01 s, ν = - 0.50 \cos 2 π \{\frac{1}{4} - \frac{1}{2}\} = 0 .$

(c) (i) Speed of the particle:
$At x = 3 cm and t = 0.01 s,$ $ν = - 0.50 \cos 2 π \{\frac{3}{4} - \frac{1}{2}\} = 0$ .
(ii) $At x = 5 cm and t = 0.01 s,$
$ν = 0 (iii) At x = 7 cm and t = 0.1 s, ν = 0 . (iv) At x = 1 cm and t = 0.011 s, ν = 50 \cos 2 π \{(\frac{1}{4}) - (\frac{0.011}{0.02})\} = - 50 \cos (\frac{3 π}{5}) = - 9.7 cm / s$
(By changing the value of t, the other two can be calculated.)

Answer:

Time taken to reach from the mean position to the extreme position, $\frac{T}{4}$ = 5 ms
Time period (T) of the wave:
$T = 4 \times 5 ms = 20 \times 10^{- 3} = 2 \times 10^{- 2} s$
Wavelength (λ) = $2 \times Distance between two mean positions$
$= 2 \times 2 cm = 4 cm$
$Frequency, f = \frac{1}{T} = \frac{1}{(2 \times 10^{- 2})} = 50 Hz Wave speed, v = λ f = 4 \times 10^{- 2} \times 50 = 200 \times 10^{- 2} = 2 m / s$

Question 12:

Time taken to reach from the mean position to the extreme position, $\frac{T}{4}$ = 5 ms
Time period (T) of the wave:
$T = 4 \times 5 ms = 20 \times 10^{- 3} = 2 \times 10^{- 2} s$
Wavelength (λ) = $2 \times Distance between two mean positions$
$= 2 \times 2 cm = 4 cm$
$Frequency, f = \frac{1}{T} = \frac{1}{(2 \times 10^{- 2})} = 50 Hz Wave speed, v = λ f = 4 \times 10^{- 2} \times 50 = 200 \times 10^{- 2} = 2 m / s$

Answer:

Given:
Wave speed, $ν = 20 cm / s$
From the graph, we can infer:
(a) Amplitude, A = 1 mm
(b) Wavelength, λ = 4 cm

(c) Wave number, $k = \frac{2 π}{λ}$
$= \frac{(2 \times 3.14)}{4} = 1.57 {cm}^{- 2}$

(d) $Time period, T = \frac{λ}{ν}$
$Frequency, f = \frac{1}{T} = \frac{v}{λ} \Rightarrow f = \frac{20}{4} = 5 Hz$

Question 13:

Given:
Wave speed, $ν = 20 cm / s$
From the graph, we can infer:
(a) Amplitude, A = 1 mm
(b) Wavelength, λ = 4 cm

(c) Wave number, $k = \frac{2 π}{λ}$
$= \frac{(2 \times 3.14)}{4} = 1.57 {cm}^{- 2}$

(d) $Time period, T = \frac{λ}{ν}$
$Frequency, f = \frac{1}{T} = \frac{v}{λ} \Rightarrow f = \frac{20}{4} = 5 Hz$

Answer:

Given,
Wave speed (v) = 10 ms⁻¹
Time period (T) = 20 ms $= 20 \times 10^{- 3} = 2 \times 10^{- 2} s$

(a) Wavelength of the wave:
$λ = ν t = 10 \times 2 \times 10^{- 2} = 0.02 m = 20 cm$

(b) Displacement of the particle at a certain instant:
$y = a \sin (ω t - k x) \Rightarrow 1.5 = a \sin (ω t - k x)$
Phase difference of the particle at a distance x = 10 cm:
$ϕ = \frac{2 π x}{λ} = 2 π \times \frac{10}{20} = π$
$The displacement is given by y' = a \sin (ω t - k x + π) = a \sin (ω t - k x) = 1.5 mm ∴ Displacement = 1.5 mm$

Question 14:

Given,
Wave speed (v) = 10 ms⁻¹
Time period (T) = 20 ms $= 20 \times 10^{- 3} = 2 \times 10^{- 2} s$

(a) Wavelength of the wave:
$λ = ν t = 10 \times 2 \times 10^{- 2} = 0.02 m = 20 cm$

(b) Displacement of the particle at a certain instant:
$y = a \sin (ω t - k x) \Rightarrow 1.5 = a \sin (ω t - k x)$
Phase difference of the particle at a distance x = 10 cm:
$ϕ = \frac{2 π x}{λ} = 2 π \times \frac{10}{20} = π$
$The displacement is given by y' = a \sin (ω t - k x + π) = a \sin (ω t - k x) = 1.5 mm ∴ Displacement = 1.5 mm$

Answer:

Given,
Length of the steel wire = 64 cm
Weight = 5 g
Applied force = 8 N
Thus, we have:
$Mass per unit length = \frac{5}{64} gm / cm Tension, T = 8 N = 8 \times 10^{5} dyn Speed, v = \sqrt{(\frac{T}{m})} = \sqrt{\frac{(8 \times 10^{5} \times 64)}{5}} = 3200 cm / s = 32 m / s$

Question 15:

Given,
Length of the steel wire = 64 cm
Weight = 5 g
Applied force = 8 N
Thus, we have:
$Mass per unit length = \frac{5}{64} gm / cm Tension, T = 8 N = 8 \times 10^{5} dyn Speed, v = \sqrt{(\frac{T}{m})} = \sqrt{\frac{(8 \times 10^{5} \times 64)}{5}} = 3200 cm / s = 32 m / s$

Answer:

Given,
Length of the string = 20 cm
Linear mass density of the string = 0.40 g cm⁻¹
Applied tension = 16 N = $16 \times 10^{5} dyn$
Velocity of the wave:
$ν = \sqrt{(\frac{T}{m})} = \sqrt{\frac{(16 \times 10^{5})}{0.4}} = 2000 cm / s$
∴ Time taken to reach the other end $= \frac{20}{2000} = 0.01 s$
Time taken to see the pulse again in the original position $= 0.01 \times 2 = 0.02 s$

(b) At t = 0.01 s, there will be a trough at the right end as it is reflected.

Question 16:

Given,
Length of the string = 20 cm
Linear mass density of the string = 0.40 g cm⁻¹
Applied tension = 16 N = $16 \times 10^{5} dyn$
Velocity of the wave:
$ν = \sqrt{(\frac{T}{m})} = \sqrt{\frac{(16 \times 10^{5})}{0.4}} = 2000 cm / s$
∴ Time taken to reach the other end $= \frac{20}{2000} = 0.01 s$
Time taken to see the pulse again in the original position $= 0.01 \times 2 = 0.02 s$

(b) At t = 0.01 s, there will be a trough at the right end as it is reflected.

Answer:

Given,
Linear mass density of the string = 0.5 gcm⁻¹
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms⁻¹

The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.
$Phase change = 0$
(a) $Total distance, S = 20 + 20 = 40 cm Wave speed, ν = 20 m / s$
Time taken to regain shape:
$Time = \frac{S}{ν} = \frac{40}{20} = 2 s$

(b) The wave regain its shape after covering a period distance $= 2 \times 30 = 60$ cm
$∴ Time period = \frac{60}{20} = 3 s$

(c) Frequency, $n = \frac{1}{Time period} = \frac{1}{3} s^{- 1}$
We know:
$n = \frac{1}{2 l} \sqrt{(\frac{T}{m})}$
Here, T is the tension in the string.
Now,

$m = Mass per unit length = 0.5 gm / cm \Rightarrow \frac{1}{3} = \frac{1}{(2 \times 30)} \sqrt{(\frac{T}{0.5})} \Rightarrow T = 400 \times 0.5 = 200 dyn = 2 \times 10^{- 3} N$

Question 17:

Given,
Linear mass density of the string = 0.5 gcm⁻¹
Total length of the string = 30 cm
Speed of the wave pulse = 20 cms⁻¹

The crest reflects the crest here because the wave is travelling from a denser medium to a rarer medium.
$Phase change = 0$
(a) $Total distance, S = 20 + 20 = 40 cm Wave speed, ν = 20 m / s$
Time taken to regain shape:
$Time = \frac{S}{ν} = \frac{40}{20} = 2 s$

(b) The wave regain its shape after covering a period distance $= 2 \times 30 = 60$ cm
$∴ Time period = \frac{60}{20} = 3 s$

(c) Frequency, $n = \frac{1}{Time period} = \frac{1}{3} s^{- 1}$
We know:
$n = \frac{1}{2 l} \sqrt{(\frac{T}{m})}$
Here, T is the tension in the string.
Now,

$m = Mass per unit length = 0.5 gm / cm \Rightarrow \frac{1}{3} = \frac{1}{(2 \times 30)} \sqrt{(\frac{T}{0.5})} \Rightarrow T = 400 \times 0.5 = 200 dyn = 2 \times 10^{- 3} N$

Answer:

Let:
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v₁.
Thus, we have:
â€‹ $ν_{1} = \sqrt{(\frac{T}{m_{1}})}$
The mass per unit length can be given as
$m_{1} = (\frac{ρ_{1} a_{1} I_{1}}{I_{1}}) = ρ_{1} a_{1} \Rightarrow ν_{1} = \sqrt{(\frac{T}{ρ_{1} a_{1}})} . . . (1)$
Let the velocity of the first string be v ₂.
Thus, we have:
$ν_{2} = \sqrt{(\frac{T}{m_{2}})} \Rightarrow ν_{2} = \sqrt{(\frac{T}{ρ_{2} a_{2}})} . . . (2)$
Given,
$ν_{1} = 2 ν_{2} \Rightarrow \sqrt{(\frac{T}{a_{1} ρ_{1}})} = 2 \sqrt{(\frac{T}{a_{2} ρ_{2}})} \Rightarrow (\frac{T}{a_{1} ρ_{1}}) = 4 (\frac{T}{a_{2} ρ_{2}}) \Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{1}{4} \Rightarrow ρ_{1} : ρ_{2} = 1 : 4$

Question 18:

Let:
m = Mass per unit length of the first wire
a = Area of the cross section
ρ = Density of the wire
T = Tension
Let the velocity of the first string be v₁.
Thus, we have:
â€‹ $ν_{1} = \sqrt{(\frac{T}{m_{1}})}$
The mass per unit length can be given as
$m_{1} = (\frac{ρ_{1} a_{1} I_{1}}{I_{1}}) = ρ_{1} a_{1} \Rightarrow ν_{1} = \sqrt{(\frac{T}{ρ_{1} a_{1}})} . . . (1)$
Let the velocity of the first string be v ₂.
Thus, we have:
$ν_{2} = \sqrt{(\frac{T}{m_{2}})} \Rightarrow ν_{2} = \sqrt{(\frac{T}{ρ_{2} a_{2}})} . . . (2)$
Given,
$ν_{1} = 2 ν_{2} \Rightarrow \sqrt{(\frac{T}{a_{1} ρ_{1}})} = 2 \sqrt{(\frac{T}{a_{2} ρ_{2}})} \Rightarrow (\frac{T}{a_{1} ρ_{1}}) = 4 (\frac{T}{a_{2} ρ_{2}}) \Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{1}{4} \Rightarrow ρ_{1} : ρ_{2} = 1 : 4$

Answer:

Given,
Wave equation, $y = (0 \cdot 02 m) \sin (1 \cdot 0 m^{- 1}) x + (30 s^{- 1}) t$

$Let : Mass per unit length, m = 1.2 \times 10^{- 4} kg / m From the wave equation, we have : k = 1 m^{- 1} = \frac{2 π}{λ} And, ω = 30 s^{- 1} = 2 π f$

Velocity of the wave in the stretched string is given by

$ν = λ f = \frac{ω}{k} = \frac{30}{1} \Rightarrow v = 30 m / s We know : v = \sqrt{\frac{T}{m}} \Rightarrow 30 = \sqrt{(\frac{T}{1.2 \times 10^{- 4}})} \Rightarrow T = 108 \times 10^{- 3} = 0.108 N$

So, the tension in the string is 0.108 N.

Question 19:

Given,
Wave equation, $y = (0 \cdot 02 m) \sin (1 \cdot 0 m^{- 1}) x + (30 s^{- 1}) t$

$Let : Mass per unit length, m = 1.2 \times 10^{- 4} kg / m From the wave equation, we have : k = 1 m^{- 1} = \frac{2 π}{λ} And, ω = 30 s^{- 1} = 2 π f$

Velocity of the wave in the stretched string is given by

$ν = λ f = \frac{ω}{k} = \frac{30}{1} \Rightarrow v = 30 m / s We know : v = \sqrt{\frac{T}{m}} \Rightarrow 30 = \sqrt{(\frac{T}{1.2 \times 10^{- 4}})} \Rightarrow T = 108 \times 10^{- 3} = 0.108 N$

So, the tension in the string is 0.108 N.

Answer:

Given,
Amplitude of the wave = 1 cm
Frequency of the wave, $f = \frac{200}{2} = 100 Hz$
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N

(a) Velocity of the wave is given by
$v = \sqrt{\frac{T}{m}}$
Thus, we have:
$v = \sqrt{(\frac{90}{0.1})} = 30 m / s Now, Wavelength, λ = \frac{v}{f} = \frac{30}{100} = 0.3 m \Rightarrow λ = 30 cm$

(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
$y = (1 cm) \cos 2 π \{(\frac{t}{0.01 s}) - (\frac{x}{30 cm})\}$ ...(1)

(c) Using $\cos (- θ) = \cos θ$ in equation (1), we get:
$y = 1 \cos 2 π (\frac{x}{30} - \frac{t}{0.01})$
$Velocity, v = \frac{d y}{d t} \Rightarrow v = (\frac{2 π}{0.01}) \sin 2 π \{\frac{x}{30} - \frac{t}{0.01}\} And, Acceleration, a = \frac{d ν}{d t} \Rightarrow a = \{\frac{4 π^{2}}{{(0.01)}^{2}}\} \cos 2 π \{(\frac{x}{30}) - (\frac{t}{0.01})\} When x = 50 cm, t = 10 ms = 10 \times 10^{- 3} s .$

Now,

$v = (\frac{2 π}{0.01}) \sin 2 π \{(\frac{5}{3}) - (\frac{0.01}{0.01})\} = (\frac{2 π}{0.01}) \sin (2 π \times \frac{2}{3}) = - (\frac{2 π}{0.01}) \sin \frac{4 π}{3} = - 200 π \sin \frac{π}{3} = - 200 π \times \frac{\sqrt{3}}{2} = - 544 cm / s = - 5.4 m / s$
In magnitude, v = 5.4 m/s.
Similarly,
$a = \{\frac{4 π^{2}}{{(0.01)}^{2}}\} \cos 2 π \{(\frac{5}{3}) - 1\} = 4 π^{2} \times 10^{4} \times \frac{1}{2} \approx 2 \times 10^{5} cm / s^{2} or 2 km / s^{2}$

Question 20:

Given,
Amplitude of the wave = 1 cm
Frequency of the wave, $f = \frac{200}{2} = 100 Hz$
Mass per unit length, m = 0.1 kg/m
Applied tension, T = 90 N

(a) Velocity of the wave is given by
$v = \sqrt{\frac{T}{m}}$
Thus, we have:
$v = \sqrt{(\frac{90}{0.1})} = 30 m / s Now, Wavelength, λ = \frac{v}{f} = \frac{30}{100} = 0.3 m \Rightarrow λ = 30 cm$

(b) At x = 0, displacement is maximum.
Thus, the wave equation is given by
$y = (1 cm) \cos 2 π \{(\frac{t}{0.01 s}) - (\frac{x}{30 cm})\}$ ...(1)

(c) Using $\cos (- θ) = \cos θ$ in equation (1), we get:
$y = 1 \cos 2 π (\frac{x}{30} - \frac{t}{0.01})$
$Velocity, v = \frac{d y}{d t} \Rightarrow v = (\frac{2 π}{0.01}) \sin 2 π \{\frac{x}{30} - \frac{t}{0.01}\} And, Acceleration, a = \frac{d ν}{d t} \Rightarrow a = \{\frac{4 π^{2}}{{(0.01)}^{2}}\} \cos 2 π \{(\frac{x}{30}) - (\frac{t}{0.01})\} When x = 50 cm, t = 10 ms = 10 \times 10^{- 3} s .$

Now,

$v = (\frac{2 π}{0.01}) \sin 2 π \{(\frac{5}{3}) - (\frac{0.01}{0.01})\} = (\frac{2 π}{0.01}) \sin (2 π \times \frac{2}{3}) = - (\frac{2 π}{0.01}) \sin \frac{4 π}{3} = - 200 π \sin \frac{π}{3} = - 200 π \times \frac{\sqrt{3}}{2} = - 544 cm / s = - 5.4 m / s$
In magnitude, v = 5.4 m/s.
Similarly,
$a = \{\frac{4 π^{2}}{{(0.01)}^{2}}\} \cos 2 π \{(\frac{5}{3}) - 1\} = 4 π^{2} \times 10^{4} \times \frac{1}{2} \approx 2 \times 10^{5} cm / s^{2} or 2 km / s^{2}$

Answer:

Given,
Length of the string, L = 40 cm
Mass of the string = 10 gm
Mass per unit length $= \frac{10}{40} = \frac{1}{4} (gm / cm)$
Spring constant, k = 160 N/m

$Deflection, x = 1 cm = 0.01 m Tension, T = k x = 160 \times 0.01 \Rightarrow T = 1.6 N = 16 \times 10^{4} dyn Now, v = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{16 \times 10^{4}}{\frac{1}{4}})} \Rightarrow v = 8 \times 10^{2} cm / s = 800 cm / s$

∴ Time taken by the pulse to reach the spring, $t = \frac{40}{800} = \frac{1}{20} = 0.05 s$

Question 21:

Given,
Length of the string, L = 40 cm
Mass of the string = 10 gm
Mass per unit length $= \frac{10}{40} = \frac{1}{4} (gm / cm)$
Spring constant, k = 160 N/m

$Deflection, x = 1 cm = 0.01 m Tension, T = k x = 160 \times 0.01 \Rightarrow T = 1.6 N = 16 \times 10^{4} dyn Now, v = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{16 \times 10^{4}}{\frac{1}{4}})} \Rightarrow v = 8 \times 10^{2} cm / s = 800 cm / s$

∴ Time taken by the pulse to reach the spring, $t = \frac{40}{800} = \frac{1}{20} = 0.05 s$

Answer:

Given,
Mass of each block = $m_{1} = m_{2} = 3.2 kg$
Linear mass density of wire AB = 10 gm⁻¹ = 0.01 kgm⁻¹
Linear mass density of wire CD = 8 gm⁻¹ = 0.008 kgm⁻¹
For string CD, velocity is defined as $v = \sqrt{(\frac{T}{m})}$ .
Here, T is the tension and m is the mass per unit length.
For string CD,
$T = 3.2 \times g$
Thus, we have:
$v = \sqrt{\frac{(3.2 \times 10)}{0.008}} = \sqrt{\frac{(32 \times 10^{3})}{8}} = 2 \times 10 \sqrt{10} = 20 \times 3.14 \approx 63 ms$

For string AB,
$T = 2 \times 3.2 g = 64 N Thus, we have : v = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{64}{0.01})} = \sqrt{6400} = 80 m / s$

Question 22:

Given,
Mass of each block = $m_{1} = m_{2} = 3.2 kg$
Linear mass density of wire AB = 10 gm⁻¹ = 0.01 kgm⁻¹
Linear mass density of wire CD = 8 gm⁻¹ = 0.008 kgm⁻¹
For string CD, velocity is defined as $v = \sqrt{(\frac{T}{m})}$ .
Here, T is the tension and m is the mass per unit length.
For string CD,
$T = 3.2 \times g$
Thus, we have:
$v = \sqrt{\frac{(3.2 \times 10)}{0.008}} = \sqrt{\frac{(32 \times 10^{3})}{8}} = 2 \times 10 \sqrt{10} = 20 \times 3.14 \approx 63 ms$

For string AB,
$T = 2 \times 3.2 g = 64 N Thus, we have : v = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{64}{0.01})} = \sqrt{6400} = 80 m / s$

Answer:

Given,
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string:
$m = \frac{4.5 \times 10^{- 3}}{2.25} = 2 \times 10^{- 3} kg / m T = 2 g = 20 N Wave speed, ν = \sqrt{(\frac{T}{m})} = \sqrt{\frac{20}{(2 \times 10^{- 3})}} = \sqrt{10^{4}} = 10^{2} m / s = 100 m / s$

Time taken by the disturbance to reach the pulley:
$t = (\frac{s}{ν}) = \frac{2}{100} = 0.02 s$

Question 23:

Given,
Mass of the block = 2 kg
Total length of the string = 2 + 0.25 = 2.25 m
Mass per unit length of the string:
$m = \frac{4.5 \times 10^{- 3}}{2.25} = 2 \times 10^{- 3} kg / m T = 2 g = 20 N Wave speed, ν = \sqrt{(\frac{T}{m})} = \sqrt{\frac{20}{(2 \times 10^{- 3})}} = \sqrt{10^{4}} = 10^{2} m / s = 100 m / s$

Time taken by the disturbance to reach the pulley:
$t = (\frac{s}{ν}) = \frac{2}{100} = 0.02 s$

Answer:

Given,
Mass of the block = 4.0 kg
Linear mass density, $m = 19.2 \times 10^{- 3} kg / m$
From the free body diagram,
$T - 4 g - 4 a = 0 \Rightarrow T = 4 (a + g) = 4 (2 + 10) = 48 N$

$Wave speed, ν = \sqrt{(\frac{T}{m})} = \sqrt{\frac{48}{19.2 \times 10^{3}}} = \sqrt{(2.5 \times 10^{- 3})} = 50 m / s$

Question 24:

Given,
Mass of the block = 4.0 kg
Linear mass density, $m = 19.2 \times 10^{- 3} kg / m$
From the free body diagram,
$T - 4 g - 4 a = 0 \Rightarrow T = 4 (a + g) = 4 (2 + 10) = 48 N$

$Wave speed, ν = \sqrt{(\frac{T}{m})} = \sqrt{\frac{48}{19.2 \times 10^{3}}} = \sqrt{(2.5 \times 10^{- 3})} = 50 m / s$

Answer:

Given,
Speed of the transverse pulse when the car is at rest, v₁ = 60 cm s⁻¹
Speed of the transverse pulse when the car accelerates, v₂= 62 cm s⁻¹
Let:
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Now,
$Wave speed, ν = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{M g}{m})} When car is at rest : Tension in the string, T = M g \Rightarrow v_{1} = \sqrt{(\frac{M g}{m})} \Rightarrow \frac{M g}{m} = {(60)}^{2} . . . (i)$
When car is having acceleration:
Tension, $T = \sqrt{{(M a)}^{2} + {(M g)}^{2}}$
$Again, ν_{2} = \sqrt{(\frac{T}{m})} \Rightarrow 62 = \frac{{[{(M a)}^{2} + {(M g)}^{2}]}^{1 / 4}}{m^{1 / 2}}$
$\Rightarrow \sqrt{\frac{[{(M a)}^{2} + {(M g)}^{2}]}{m}} = {(62)}^{2} . . . (ii)$
From equations (i) and (ii), we get:
$(\frac{M g}{m}) \times \frac{m}{\sqrt{{(M a)}^{2} + {(M g)}^{2}}} = {(\frac{60}{62})}^{2} \Rightarrow \frac{g}{\sqrt{(a^{2} + g^{2})}} = 0.936 \Rightarrow \frac{g^{2}}{(a^{2} + g^{2})} = 0.876 \Rightarrow (a^{2} + 100) 0.876 = 100 \Rightarrow a^{2} = \frac{12.4}{0.867} = 14.15 \Rightarrow a = 3.76 m / s^{2} Therefore, acceleration of the car is 3.76 m / s^{2} .$

Question 25:

Given,
Speed of the transverse pulse when the car is at rest, v₁ = 60 cm s⁻¹
Speed of the transverse pulse when the car accelerates, v₂= 62 cm s⁻¹
Let:
Mass of the heavy ball suspended from the ceiling = M
Mass per unit length = m
Now,
$Wave speed, ν = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{M g}{m})} When car is at rest : Tension in the string, T = M g \Rightarrow v_{1} = \sqrt{(\frac{M g}{m})} \Rightarrow \frac{M g}{m} = {(60)}^{2} . . . (i)$
When car is having acceleration:
Tension, $T = \sqrt{{(M a)}^{2} + {(M g)}^{2}}$
$Again, ν_{2} = \sqrt{(\frac{T}{m})} \Rightarrow 62 = \frac{{[{(M a)}^{2} + {(M g)}^{2}]}^{1 / 4}}{m^{1 / 2}}$
$\Rightarrow \sqrt{\frac{[{(M a)}^{2} + {(M g)}^{2}]}{m}} = {(62)}^{2} . . . (ii)$
From equations (i) and (ii), we get:
$(\frac{M g}{m}) \times \frac{m}{\sqrt{{(M a)}^{2} + {(M g)}^{2}}} = {(\frac{60}{62})}^{2} \Rightarrow \frac{g}{\sqrt{(a^{2} + g^{2})}} = 0.936 \Rightarrow \frac{g^{2}}{(a^{2} + g^{2})} = 0.876 \Rightarrow (a^{2} + 100) 0.876 = 100 \Rightarrow a^{2} = \frac{12.4}{0.867} = 14.15 \Rightarrow a = 3.76 m / s^{2} Therefore, acceleration of the car is 3.76 m / s^{2} .$

Answer:

Let,
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity

Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part dθ at angle θ.
So,
Length of the element $= R d θ, mass = m R d θ$
Centrifugal force experienced by the element $= (m R d θ) ω^{2} R$
Resolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as
$d F = 2 m R^{2} d θ ω^{2} sinθ$
$Total force, F = \int_{0}^{π / 2} 2 m R^{2} ω^{2} \sin θ d θ = 2 m R^{2} ω^{2} [- \cos θ] = 2 m R^{2} ω^{2} And, 2 T = 2 m R^{2} ω^{2} \Rightarrow T = m R^{2} ω^{2}$
Velocity of the transverse vibration is given as
$V' = \sqrt{(\frac{T}{m})} V' = \sqrt{(\frac{m R^{2} ω^{2}}{m})} = ω R$
Linear velocity of the string, V = $ω R$
∴ Speed of the disturbance, V' = V

Question 26:

Let,
V = Linear velocity of the string
m = Mass per unit length of the the string.
R = Radius of the loop
ω = Angular velocity

Consider one half of the string, as shown in the figure.
The half loop experiences centrifugal force at every point (away from the centre) balanced by tension 2T.
Consider an element of angular part dθ at angle θ.
So,
Length of the element $= R d θ, mass = m R d θ$
Centrifugal force experienced by the element $= (m R d θ) ω^{2} R$
Resolving the centrifugal force into rectangular components,
Since the horizontal components cancel each other, the net force on the two symmetric elements is given as
$d F = 2 m R^{2} d θ ω^{2} sinθ$
$Total force, F = \int_{0}^{π / 2} 2 m R^{2} ω^{2} \sin θ d θ = 2 m R^{2} ω^{2} [- \cos θ] = 2 m R^{2} ω^{2} And, 2 T = 2 m R^{2} ω^{2} \Rightarrow T = m R^{2} ω^{2}$
Velocity of the transverse vibration is given as
$V' = \sqrt{(\frac{T}{m})} V' = \sqrt{(\frac{m R^{2} ω^{2}}{m})} = ω R$
Linear velocity of the string, V = $ω R$
∴ Speed of the disturbance, V' = V

Answer:

(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as
$v = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{m g x}{m})} \Rightarrow v = \sqrt{(g x)}$

(b) Let the time taken be dt for the small displacement dx.
Thus, we have:

$d t = \frac{d x}{v} = \frac{d x}{\sqrt{(g x)}}$
$∴ Total time, T = \int_{0}^{L} \frac{d x}{\sqrt{(g x)}} = \sqrt{(\frac{4 L}{g})}$

(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.

Now,
$t = \int_{0}^{y} \frac{d x}{\sqrt{(g x)}} = \sqrt{(\frac{4 y}{g})}$

∴ Distance travelled by the particle in this time, S = $L - y$
Using the equation of motion, we get:
$S = u t + \frac{1}{2} g t^{2} \Rightarrow L - y = (\frac{1}{2}) g \times \{{(\sqrt{\frac{4 y}{g}})}^{2}\} \Rightarrow L - y = 2 y \Rightarrow 3 y = L \Rightarrow y = \frac{L}{3}$
Thus, the particle will meet the pulse at a distance $\frac{L}{3}$ from the lower end.

Question 27:

(a) Let m be the mass per unit length of the string.
Consider an element at a distance x from the lower end.
Here,
Weight acting downwards = (mx)g
∴ Tension in the string at the upper part = mgx
The velocity of transverse vibration is given as
$v = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{m g x}{m})} \Rightarrow v = \sqrt{(g x)}$

(b) Let the time taken be dt for the small displacement dx.
Thus, we have:

$d t = \frac{d x}{v} = \frac{d x}{\sqrt{(g x)}}$
$∴ Total time, T = \int_{0}^{L} \frac{d x}{\sqrt{(g x)}} = \sqrt{(\frac{4 L}{g})}$

(c) Suppose after time t, the pulse meets the particle at a distance y from the lower end of the rope.

Now,
$t = \int_{0}^{y} \frac{d x}{\sqrt{(g x)}} = \sqrt{(\frac{4 y}{g})}$

∴ Distance travelled by the particle in this time, S = $L - y$
Using the equation of motion, we get:
$S = u t + \frac{1}{2} g t^{2} \Rightarrow L - y = (\frac{1}{2}) g \times \{{(\sqrt{\frac{4 y}{g}})}^{2}\} \Rightarrow L - y = 2 y \Rightarrow 3 y = L \Rightarrow y = \frac{L}{3}$
Thus, the particle will meet the pulse at a distance $\frac{L}{3}$ from the lower end.

Answer:

Given,
Linear density of each of two long strings A and B, m = $1.2 \times 10^{- 2} kg / m$
String A is stretched by tension T_a= 4.8 N.
String B is stretched by tension T_b= 7.5 N.
Let v_a and v_b be the speeds of the waves in strings A and B.
Now,
$v_{a} = \sqrt{\frac{T_{a}}{m}} \Rightarrow v_{a} = \sqrt{\frac{(4.8)}{(1.2 \times 10^{- 2})}} = 20 m / s v_{b} = \sqrt{\frac{T_{b}}{m}} \Rightarrow v_{b} = \sqrt{\frac{7.5}{(1.2 \times 10^{- 2})}} = 25 m / s t_{1} = 0 in string A t_{2} = 0 + 20 ms = 20 \times 10^{- 3} = 0.02 s$
Distance travelled by the wave in 0.02 s in string A:
s $= 20 \times 0.02 = 0.4 m$
Relative speed between the wave in string A and the wave in string B, v' $= 25 - 20 = 5 m / s$
Time taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m
$t' = \frac{s}{v'} = \frac{0.4}{5} = 0.08 s$

Question 28:

Given,
Linear density of each of two long strings A and B, m = $1.2 \times 10^{- 2} kg / m$
String A is stretched by tension T_a= 4.8 N.
String B is stretched by tension T_b= 7.5 N.
Let v_a and v_b be the speeds of the waves in strings A and B.
Now,
$v_{a} = \sqrt{\frac{T_{a}}{m}} \Rightarrow v_{a} = \sqrt{\frac{(4.8)}{(1.2 \times 10^{- 2})}} = 20 m / s v_{b} = \sqrt{\frac{T_{b}}{m}} \Rightarrow v_{b} = \sqrt{\frac{7.5}{(1.2 \times 10^{- 2})}} = 25 m / s t_{1} = 0 in string A t_{2} = 0 + 20 ms = 20 \times 10^{- 3} = 0.02 s$
Distance travelled by the wave in 0.02 s in string A:
s $= 20 \times 0.02 = 0.4 m$
Relative speed between the wave in string A and the wave in string B, v' $= 25 - 20 = 5 m / s$
Time taken by the wave in string B to overtake the wave in string A = Time taken by the wave in string B to cover 0.4 m
$t' = \frac{s}{v'} = \frac{0.4}{5} = 0.08 s$

Answer:

Given,
Amplitude of the transverse wave, r = 0.5 mm $= 0.5 \times 10^{- 3} m$
Frequency, f = 100 Hz
Tension, T = 100 N
Wave speed, v = 100 m/s
Thus, we have:
$ν = \sqrt{(\frac{T}{m})} \Rightarrow ν^{2} = (\frac{T}{m}) \Rightarrow m = \frac{T}{ν^{2}} = \frac{100}{{(100)}^{2}} = 0.01 kg / m Average power of the source : P_{avg} = 2 π^{2} m ν r^{2} f^{2} = 2 {(3.14)}^{2} (0.01) \times 100 \times {(0.5 \times 10^{- 3})}^{2} \times (100) = 2 \times 9.86 \times 0.25 \times 10^{- 6} \times 10^{4} = 19.7 \times 0.0025 = 0.049 W = 49 \times 10^{- 3} W = 49 mW$

Question 29:

Given,
Amplitude of the transverse wave, r = 0.5 mm $= 0.5 \times 10^{- 3} m$
Frequency, f = 100 Hz
Tension, T = 100 N
Wave speed, v = 100 m/s
Thus, we have:
$ν = \sqrt{(\frac{T}{m})} \Rightarrow ν^{2} = (\frac{T}{m}) \Rightarrow m = \frac{T}{ν^{2}} = \frac{100}{{(100)}^{2}} = 0.01 kg / m Average power of the source : P_{avg} = 2 π^{2} m ν r^{2} f^{2} = 2 {(3.14)}^{2} (0.01) \times 100 \times {(0.5 \times 10^{- 3})}^{2} \times (100) = 2 \times 9.86 \times 0.25 \times 10^{- 6} \times 10^{4} = 19.7 \times 0.0025 = 0.049 W = 49 \times 10^{- 3} W = 49 mW$

Answer:

Given,
Frequency of the wave, f = 200 Hz
Amplitude, A = 1 mm = 10⁻³ m
Linear mass density, m = 6 gm⁻³
Applied tension, T = 60 N
Now,
Let the velocity of the wave be v.
Thus, we have:
$v = \sqrt{(\frac{T}{m})} = \sqrt{\frac{(60)}{(6 \times 10^{- 3})}} = 10^{2} = 100 m / s$

(a) Average power is given as
$P_{a v e r a g e} = 2 π^{2} m ν A^{2} f^{2} = 2 \times {(3.14)}^{2} \times (6 \times 10^{- 3}) \times 100 \times (10^{- 3}) \times 200^{2} = 473 \times 10^{- 3} = 0.47 W$

(b) Length of the string = 2 m
Time required to cover this distance:
$t = \frac{2}{100} = 0.02 s Energy = Power \times t = 0.47 \times 0.02 = 9.4 \times 10^{- 3} J = 9.4 mJ$

Question 30:

Given,
Frequency of the wave, f = 200 Hz
Amplitude, A = 1 mm = 10⁻³ m
Linear mass density, m = 6 gm⁻³
Applied tension, T = 60 N
Now,
Let the velocity of the wave be v.
Thus, we have:
$v = \sqrt{(\frac{T}{m})} = \sqrt{\frac{(60)}{(6 \times 10^{- 3})}} = 10^{2} = 100 m / s$

(a) Average power is given as
$P_{a v e r a g e} = 2 π^{2} m ν A^{2} f^{2} = 2 \times {(3.14)}^{2} \times (6 \times 10^{- 3}) \times 100 \times (10^{- 3}) \times 200^{2} = 473 \times 10^{- 3} = 0.47 W$

(b) Length of the string = 2 m
Time required to cover this distance:
$t = \frac{2}{100} = 0.02 s Energy = Power \times t = 0.47 \times 0.02 = 9.4 \times 10^{- 3} J = 9.4 mJ$

Answer:

Given,
Frequency of the tuning fork, f = 440 Hz
Linear mass density, m = 0.01 kgm⁻¹
Applied tension, T = 49 N
Amplitude of the transverse wave produce by the fork = 0.50 mm
Let the wavelength of the wave be $λ$ .

(a) The speed of the transverse wave is given by
$ν = \sqrt{(\frac{T}{m})}$
$\Rightarrow v = \sqrt{\frac{49}{0.01}} = 70 m / s$
$Also, ν = \frac{f}{λ} ∴ λ = \frac{f}{v} = \frac{70}{440} = 16 cm$

(b) Maximum speed (v_max) and maximum acceleration (a_max):
$We have : y = A \sin (ω t - k x)$
$∴ ν = \frac{d y}{d t} = A ω \cos (ω t - k x) Now, ν_{\max} = (\frac{d y}{d t}) = A ω = 0.50 \times 10^{- 3} \times 2 π \times 440 = 1.3816 m / s . And, a = \frac{d^{2} y}{d t^{2}} \Rightarrow a = - A ω^{2} \sin (ω t - k x) a_{\max} = - A ω^{2} = 0.50 \times 10^{- 3} \times 4 π^{2} {(440)}^{2} = 3.8 km / s^{2}$

(c) Average rate (p) is given by
$p = 2 π^{2} ν A^{2} f^{2} = 2 \times 10 \times 0.01 \times 70 \times {(0.5 \times 10^{- 3})}^{2} \times {(440)}^{2} = 0.67 W$

Question 31:

Given,
Frequency of the tuning fork, f = 440 Hz
Linear mass density, m = 0.01 kgm⁻¹
Applied tension, T = 49 N
Amplitude of the transverse wave produce by the fork = 0.50 mm
Let the wavelength of the wave be $λ$ .

(a) The speed of the transverse wave is given by
$ν = \sqrt{(\frac{T}{m})}$
$\Rightarrow v = \sqrt{\frac{49}{0.01}} = 70 m / s$
$Also, ν = \frac{f}{λ} ∴ λ = \frac{f}{v} = \frac{70}{440} = 16 cm$

(b) Maximum speed (v_max) and maximum acceleration (a_max):
$We have : y = A \sin (ω t - k x)$
$∴ ν = \frac{d y}{d t} = A ω \cos (ω t - k x) Now, ν_{\max} = (\frac{d y}{d t}) = A ω = 0.50 \times 10^{- 3} \times 2 π \times 440 = 1.3816 m / s . And, a = \frac{d^{2} y}{d t^{2}} \Rightarrow a = - A ω^{2} \sin (ω t - k x) a_{\max} = - A ω^{2} = 0.50 \times 10^{- 3} \times 4 π^{2} {(440)}^{2} = 3.8 km / s^{2}$

(c) Average rate (p) is given by
$p = 2 π^{2} ν A^{2} f^{2} = 2 \times 10 \times 0.01 \times 70 \times {(0.5 \times 10^{- 3})}^{2} \times {(440)}^{2} = 0.67 W$

Answer:

Given,
Phase difference between the two waves travelling in the same direction, $ϕ = 90 ° = \frac{π}{2}$ .
Frequency f and wavelength $λ$ are the same. Therefore, ω will be the same.
Let the wave equations of two waves be:
$y_{1} = r \sin ω t$
$y_{2} = r \sin (ω t + \frac{π}{2})$
Here, r is the amplitude.
From the principle of superposition, we get:
$y = y_{1} + y_{2} = r \sin ω t + r \sin (ω t + \frac{π}{2}) = r [\sin ω t + \sin (ω t + \frac{π}{2})] = r [2 \sin \{(\frac{ω t + ω t + \frac{π}{2}}{2})\} \cos \{(\frac{ω t - ω t - \frac{π}{2}}{2})\}] = 2 r \sin (ω t + \frac{π}{4}) \cos (- \frac{π}{4}) = \sqrt{2} r \sin (ω t + (\frac{π}{4}))$
∴ Resultant amplitude, $r' = \sqrt{2} r = 4 \sqrt{2} mm$

Question 32:

Given,
Phase difference between the two waves travelling in the same direction, $ϕ = 90 ° = \frac{π}{2}$ .
Frequency f and wavelength $λ$ are the same. Therefore, ω will be the same.
Let the wave equations of two waves be:
$y_{1} = r \sin ω t$
$y_{2} = r \sin (ω t + \frac{π}{2})$
Here, r is the amplitude.
From the principle of superposition, we get:
$y = y_{1} + y_{2} = r \sin ω t + r \sin (ω t + \frac{π}{2}) = r [\sin ω t + \sin (ω t + \frac{π}{2})] = r [2 \sin \{(\frac{ω t + ω t + \frac{π}{2}}{2})\} \cos \{(\frac{ω t - ω t - \frac{π}{2}}{2})\}] = 2 r \sin (ω t + \frac{π}{4}) \cos (- \frac{π}{4}) = \sqrt{2} r \sin (ω t + (\frac{π}{4}))$
∴ Resultant amplitude, $r' = \sqrt{2} r = 4 \sqrt{2} mm$

Answer:

Given,
Speed of the wave pulse travelling in the opposite direction, v = 50 cm s⁻¹ = 500 mm s⁻¹
Distances travelled by the pulses:
Using s = vt, we get:
$In t = 4 ms = 4 \times 10^{- 3} s, s = ν t = 500 \times 4 \times 10^{- 3} = 2 mm . In t = 6 ms = 6 \times 10^{- 3} s, s = 500 \times 6 \times 10^{- 3} = 3 mm . In t = 8 ms = 8 \times 10^{- 3} s, s = ν t = 500 \times 8 \times 10^{- 3} = 4 mm . In t = 12 ms = 12 \times 10^{- 3} s, s = 500 \times 12 \times 10^{- 3} = 6 mm .$

The shapes of the string at different times are shown in the figure.

Question 33:

Given,
Speed of the wave pulse travelling in the opposite direction, v = 50 cm s⁻¹ = 500 mm s⁻¹
Distances travelled by the pulses:
Using s = vt, we get:
$In t = 4 ms = 4 \times 10^{- 3} s, s = ν t = 500 \times 4 \times 10^{- 3} = 2 mm . In t = 6 ms = 6 \times 10^{- 3} s, s = 500 \times 6 \times 10^{- 3} = 3 mm . In t = 8 ms = 8 \times 10^{- 3} s, s = ν t = 500 \times 8 \times 10^{- 3} = 4 mm . In t = 12 ms = 12 \times 10^{- 3} s, s = 500 \times 12 \times 10^{- 3} = 6 mm .$

The shapes of the string at different times are shown in the figure.

Answer:

Given:
Two waves have same frequency (f), which is 100 Hz.
Wavelength (λ) = 2.0 cm $= 2 \times 10^{- 2} m$
$Wave speed, ν = f \times λ = 100 \times 2 \times 10^{- 2} m / s = 2 m / s$

(a) First wave will travel the distance in 0.015 s.
$\Rightarrow x = 0.015 \times 2 = 0.03 m$
This will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:
$ϕ = \frac{2 π x}{λ} = \{\frac{2 π}{(2 \times 10^{- 2})}\} \times 0.03 = 3 π$

(b) Path difference between the two waves, x = 4 cm = 0.04 m
So, the corresponding phase difference will be as follows:
$\Rightarrow ϕ = \frac{2 π x}{λ} = \{\frac{2 π}{(2 \times 10^{- 2})} \times 0.04\} = 4 π$

(c) The waves have same frequency, same wavelength and same amplitude.
Let the wave equation for the two waves be as follows:
$y_{1} = r \sin ω t And, y_{2} = r \sin (ω t + ϕ) By using the principle of superposition : y = y_{1} + y_{2} = r [\sin ω t + (ω t + ϕ)] = 2 r \sin (ω t + \frac{ϕ}{2}) \cos (\frac{ϕ}{2})$
∴ Resultant amplitude $= 2 r \cos \frac{ϕ}{2}$
$So, when ϕ = 3 x : \Rightarrow r = 2 \times 10^{- 3} m R_{resultant} = 2 \times (2 \times 10^{- 3}) \cos (\frac{3 π}{2}) = 0$
$Again, when ϕ = 4 π : R_{resultant} = 2 \times (2 \times 10^{- 3}) \cos (\frac{4 π}{2}) = 4 \times 10^{- 3} \times 1 = 4 mm$

Question 34:

Given:
Two waves have same frequency (f), which is 100 Hz.
Wavelength (λ) = 2.0 cm $= 2 \times 10^{- 2} m$
$Wave speed, ν = f \times λ = 100 \times 2 \times 10^{- 2} m / s = 2 m / s$

(a) First wave will travel the distance in 0.015 s.
$\Rightarrow x = 0.015 \times 2 = 0.03 m$
This will be the path difference between the two waves.
So, the corresponding phase difference will be as follows:
$ϕ = \frac{2 π x}{λ} = \{\frac{2 π}{(2 \times 10^{- 2})}\} \times 0.03 = 3 π$

(b) Path difference between the two waves, x = 4 cm = 0.04 m
So, the corresponding phase difference will be as follows:
$\Rightarrow ϕ = \frac{2 π x}{λ} = \{\frac{2 π}{(2 \times 10^{- 2})} \times 0.04\} = 4 π$

(c) The waves have same frequency, same wavelength and same amplitude.
Let the wave equation for the two waves be as follows:
$y_{1} = r \sin ω t And, y_{2} = r \sin (ω t + ϕ) By using the principle of superposition : y = y_{1} + y_{2} = r [\sin ω t + (ω t + ϕ)] = 2 r \sin (ω t + \frac{ϕ}{2}) \cos (\frac{ϕ}{2})$
∴ Resultant amplitude $= 2 r \cos \frac{ϕ}{2}$
$So, when ϕ = 3 x : \Rightarrow r = 2 \times 10^{- 3} m R_{resultant} = 2 \times (2 \times 10^{- 3}) \cos (\frac{3 π}{2}) = 0$
$Again, when ϕ = 4 π : R_{resultant} = 2 \times (2 \times 10^{- 3}) \cos (\frac{4 π}{2}) = 4 \times 10^{- 3} \times 1 = 4 mm$

Answer:

Length of a stretched string (L) = 1 m
Wave speed (v) = 60 m/s
Fundamental frequency (f₀) of vibration is given as follows:
$f_{0} = \frac{v}{2 L}$
$= \frac{60}{2} \times 1 = 30 s^{- 1} = 30 Hz$

Question 35:

Length of a stretched string (L) = 1 m
Wave speed (v) = 60 m/s
Fundamental frequency (f₀) of vibration is given as follows:
$f_{0} = \frac{v}{2 L}$
$= \frac{60}{2} \times 1 = 30 s^{- 1} = 30 Hz$

Answer:

Given:
Length of the wire (L)= 2.00 m
Fundamental frequency of the vibration (f₀) = 100 Hz
Applied tension (T) = 160 N

$Fundamental frequency, f_{0} = \frac{1}{2 L} \sqrt{(\frac{T}{m})} \Rightarrow 10 = \frac{1}{4} \sqrt{\frac{160}{m}} \Rightarrow m = 1 \times 10^{- 3} kg / m \Rightarrow m = 1 g / m$
So, the linear mass density of the wire is 1 g/m.

Question 36:

Given:
Length of the wire (L)= 2.00 m
Fundamental frequency of the vibration (f₀) = 100 Hz
Applied tension (T) = 160 N

$Fundamental frequency, f_{0} = \frac{1}{2 L} \sqrt{(\frac{T}{m})} \Rightarrow 10 = \frac{1}{4} \sqrt{\frac{160}{m}} \Rightarrow m = 1 \times 10^{- 3} kg / m \Rightarrow m = 1 g / m$
So, the linear mass density of the wire is 1 g/m.

Answer:

Given:
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m) $= (\frac{4}{80}) g / cm = 0.005 kg / m$

$Wave s peed, ν = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{50}{0.005})} = 100 m / s$

$Fundamental f requency, f_{o} = \frac{1}{2 L} \sqrt{(\frac{T}{m})} = \frac{1}{2 \times 0.8} \times \sqrt{(\frac{50}{0.005})} = \frac{100}{2 \times 0.8} = 62.5 Hz First harmonic = 62.5 Hz If f_{4} = frequency of the fourth harmonic : \Rightarrow f_{4} = 4 f_{0} = 62.5 \times 4 \Rightarrow f_{4} = 250 Hz Wavelength of the fourth harmonic, λ_{4} = \frac{ν}{f_{4}} = \frac{100}{250} \Rightarrow λ_{4} = 0.4 m = 40 cm$

Question 37:

Given:
Mass of the steel wire = 4.0 g
Length of the steel wire = 80 cm = 0.80 m
Tension in the wire = 50 N
Linear mass density (m) $= (\frac{4}{80}) g / cm = 0.005 kg / m$

$Wave s peed, ν = \sqrt{(\frac{T}{m})} = \sqrt{(\frac{50}{0.005})} = 100 m / s$

$Fundamental f requency, f_{o} = \frac{1}{2 L} \sqrt{(\frac{T}{m})} = \frac{1}{2 \times 0.8} \times \sqrt{(\frac{50}{0.005})} = \frac{100}{2 \times 0.8} = 62.5 Hz First harmonic = 62.5 Hz If f_{4} = frequency of the fourth harmonic : \Rightarrow f_{4} = 4 f_{0} = 62.5 \times 4 \Rightarrow f_{4} = 250 Hz Wavelength of the fourth harmonic, λ_{4} = \frac{ν}{f_{4}} = \frac{100}{250} \Rightarrow λ_{4} = 0.4 m = 40 cm$

Answer:

Given:
Length of the piano wire (L)= 90.0 cm = 0.90 m
Mass of the wire = 6.00 g = 0.006 kg
Fundamental frequency (f_o) = 261.63 Hz

$Linear mass density, m = (\frac{6}{90}) gm / cm = \{\frac{6 \times 10^{- 3}}{90 \times 10^{- 2}}\} kg / m = (\frac{6}{900}) kg / m$

$Fundamental f requency, f_{o} = \frac{1}{2 L} \sqrt{(\frac{T}{m})}$
$\Rightarrow 261.63 = \frac{1}{(2 \times 0.09)} \sqrt{\{\frac{T \times 900}{6}\}} \Rightarrow 0.18 \times 261.63 = \sqrt{150 T} \Rightarrow 150 T = {(261.63 \times 0.18)}^{2} \Rightarrow T = \frac{{(261.63 \times 0.18)}^{2}}{150} = 1478.52 N \approx 1480 N$

Hence, the tension in the piano wire is 1480 N.

Question 38:

Given:
Length of the piano wire (L)= 90.0 cm = 0.90 m
Mass of the wire = 6.00 g = 0.006 kg
Fundamental frequency (f_o) = 261.63 Hz

$Linear mass density, m = (\frac{6}{90}) gm / cm = \{\frac{6 \times 10^{- 3}}{90 \times 10^{- 2}}\} kg / m = (\frac{6}{900}) kg / m$

$Fundamental f requency, f_{o} = \frac{1}{2 L} \sqrt{(\frac{T}{m})}$
$\Rightarrow 261.63 = \frac{1}{(2 \times 0.09)} \sqrt{\{\frac{T \times 900}{6}\}} \Rightarrow 0.18 \times 261.63 = \sqrt{150 T} \Rightarrow 150 T = {(261.63 \times 0.18)}^{2} \Rightarrow T = \frac{{(261.63 \times 0.18)}^{2}}{150} = 1478.52 N \approx 1480 N$

Hence, the tension in the piano wire is 1480 N.

Answer:

Given:
Length of the sonometer wire (L) = 1.50 m
Let the $f irst harmonic be f_{0} and the second harmonic be f_{1} .$
According to the question, $f_{1} = 256 Hz$
$1st harmonic for fundamental frequency, f_{0} = \frac{f_{1}}{2} = \frac{256}{2} = 128 Hz$

When the fundamental wave is produced, we have:
$\frac{λ}{2} = 1.5 m \Rightarrow λ = 3 m$
$W ave speed, v = f_{0} λ \Rightarrow v = 128 \times 3 = 384 m / s$

Question 39:

Given:
Length of the sonometer wire (L) = 1.50 m
Let the $f irst harmonic be f_{0} and the second harmonic be f_{1} .$
According to the question, $f_{1} = 256 Hz$
$1st harmonic for fundamental frequency, f_{0} = \frac{f_{1}}{2} = \frac{256}{2} = 128 Hz$

When the fundamental wave is produced, we have:
$\frac{λ}{2} = 1.5 m \Rightarrow λ = 3 m$
$W ave speed, v = f_{0} λ \Rightarrow v = 128 \times 3 = 384 m / s$

Answer:

Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm

$Mass per unit length, m = \frac{12}{1.5} g / m = 8 \times 10^{- 3} kg / m$

$Tension in the wire, T = 9 \times g = 90 N$

Fundamental frequency is given by:
$f_{0} = \frac{1}{2 L} \sqrt{(\frac{T}{m})}$
For second harmonic (when two loops are produced):
$f_{1} = 2 f_{0} = \frac{1}{1.5} \sqrt{(\frac{90}{8} \times 10^{- 3})} = \frac{(106.06)}{1.5} = 70.7 Hz \approx 70 Hz$

Question 40:

Given:
Length of the wire between two pulleys (L) = 1.5 m
Mass of the wire = 12 gm

$Mass per unit length, m = \frac{12}{1.5} g / m = 8 \times 10^{- 3} kg / m$

$Tension in the wire, T = 9 \times g = 90 N$

Fundamental frequency is given by:
$f_{0} = \frac{1}{2 L} \sqrt{(\frac{T}{m})}$
For second harmonic (when two loops are produced):
$f_{1} = 2 f_{0} = \frac{1}{1.5} \sqrt{(\frac{90}{8} \times 10^{- 3})} = \frac{(106.06)}{1.5} = 70.7 Hz \approx 70 Hz$

Answer:

Given:
Length of the stretched string (L) = 1.00 m
Mass of the string =40 g
String is attached to the tuning fork that vibrates at the frequency (f) = 128 Hz
Linear mass density (m) $= (40 \times 10^{- 3}) kg / m$
No. of loops formed, (n) = 4
$L = \frac{n λ}{2} \Rightarrow λ = \frac{2 L}{n} = \frac{2 \times 1}{4} \Rightarrow λ = 0.5 m W ave speed (v) = f λ = 128 \times 0.5 \Rightarrow v = 64 m / s$

$We know : v = \sqrt{(\frac{T}{m})} \Rightarrow T = ν^{2} m = {(64)}^{2} \times 40 \times 10^{- 3} = 163.84 \approx 164 N$
Hence, the tension in the string if it is to vibrate in four loops is 164 N.

Question 41:

Given:
Length of the stretched string (L) = 1.00 m
Mass of the string =40 g
String is attached to the tuning fork that vibrates at the frequency (f) = 128 Hz
Linear mass density (m) $= (40 \times 10^{- 3}) kg / m$
No. of loops formed, (n) = 4
$L = \frac{n λ}{2} \Rightarrow λ = \frac{2 L}{n} = \frac{2 \times 1}{4} \Rightarrow λ = 0.5 m W ave speed (v) = f λ = 128 \times 0.5 \Rightarrow v = 64 m / s$

$We know : v = \sqrt{(\frac{T}{m})} \Rightarrow T = ν^{2} m = {(64)}^{2} \times 40 \times 10^{- 3} = 163.84 \approx 164 N$
Hence, the tension in the string if it is to vibrate in four loops is 164 N.

Answer:

Given:
Wire makes a resonant frequency of 240 Hz and 320 Hz when its both ends are fixed.
Therefore, fundamental frequency (f₀) of the wire must be the factor of 240 Hz and 320 Hz.
(a) Maximum value of fundamental frequency, f₀₌ 80 Hz

(b) Wave speed (v) = 40 m/s
And if $λ is the wavelength :$
$\frac{λ}{2} = L$

$∴ v = λ \times f_{0} \Rightarrow v = 2 \times L \times f_{0} \Rightarrow L = \frac{40}{2 \times 80} \Rightarrow L = \frac{1}{4} m = 0.25 m$

Question 42:

Given:
Wire makes a resonant frequency of 240 Hz and 320 Hz when its both ends are fixed.
Therefore, fundamental frequency (f₀) of the wire must be the factor of 240 Hz and 320 Hz.
(a) Maximum value of fundamental frequency, f₀₌ 80 Hz

(b) Wave speed (v) = 40 m/s
And if $λ is the wavelength :$
$\frac{λ}{2} = L$

$∴ v = λ \times f_{0} \Rightarrow v = 2 \times L \times f_{0} \Rightarrow L = \frac{40}{2 \times 80} \Rightarrow L = \frac{1}{4} m = 0.25 m$

Answer:

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be 'n' loops and $λ$ be the wavelength.
∴ $λ$ = $2 \times S eparation between the consecutive nodes$

$λ_{1} = 2 \times 2 = 4 cm$

$λ_{2} = 2 \times 1.6 = 3.2 cm$
Length of the wire is L.

In the first case:
$L = (\frac{n λ_{1}}{2})$
In the second case:
$L = (n + 1) \frac{λ_{2}}{2} \Rightarrow \frac{n λ_{1}}{2} = (n + 1) \frac{λ_{2}}{2} \Rightarrow n \times 4 = (n + 1) (3.2) \Rightarrow 4 n - 3.2 n = 3.2 \Rightarrow 0.8 n = 3.2 \Rightarrow n = 4$
∴ Length of the string, $L = \frac{(n λ_{1})}{2} = \frac{(4 \times 4)}{2} = 8 cm$

Question 43:

Given:
Separation between two consecutive nodes when the string vibrates in resonant mode = 2.0 cm
Let there be 'n' loops and $λ$ be the wavelength.
∴ $λ$ = $2 \times S eparation between the consecutive nodes$

$λ_{1} = 2 \times 2 = 4 cm$

$λ_{2} = 2 \times 1.6 = 3.2 cm$
Length of the wire is L.

In the first case:
$L = (\frac{n λ_{1}}{2})$
In the second case:
$L = (n + 1) \frac{λ_{2}}{2} \Rightarrow \frac{n λ_{1}}{2} = (n + 1) \frac{λ_{2}}{2} \Rightarrow n \times 4 = (n + 1) (3.2) \Rightarrow 4 n - 3.2 n = 3.2 \Rightarrow 0.8 n = 3.2 \Rightarrow n = 4$
∴ Length of the string, $L = \frac{(n λ_{1})}{2} = \frac{(4 \times 4)}{2} = 8 cm$

Answer:

Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s

$Wavelength, λ = \frac{v}{f} = \frac{220}{660} = \frac{1}{3} m$
(a) No. of loops, n = 3
∴ $L = \frac{n}{2} λ$
$\Rightarrow L = \frac{3}{2} \times \frac{1}{3} \Rightarrow L = \frac{1}{2} m = 50 cm$

(b) Equation of resultant stationary wave can be given by:
$y = 2 A \cos (\frac{2 π x}{λ}) \sin (\frac{2 π v L}{λ}) \Rightarrow y = 0.5 \cos (\frac{2 π x}{\frac{1}{3}}) \sin (\frac{2 π \times 220 \times t}{\frac{1}{3}}) \Rightarrow y = 0.5 cm \cos (6 π x m^{- 1}) \sin (1320 π t s^{- 1})$

Question 44:

Given:
Frequency (f) = 660 Hz
Wave speed (v) = 220 m/s

$Wavelength, λ = \frac{v}{f} = \frac{220}{660} = \frac{1}{3} m$
(a) No. of loops, n = 3
∴ $L = \frac{n}{2} λ$
$\Rightarrow L = \frac{3}{2} \times \frac{1}{3} \Rightarrow L = \frac{1}{2} m = 50 cm$

(b) Equation of resultant stationary wave can be given by:
$y = 2 A \cos (\frac{2 π x}{λ}) \sin (\frac{2 π v L}{λ}) \Rightarrow y = 0.5 \cos (\frac{2 π x}{\frac{1}{3}}) \sin (\frac{2 π \times 220 \times t}{\frac{1}{3}}) \Rightarrow y = 0.5 cm \cos (6 π x m^{- 1}) \sin (1320 π t s^{- 1})$

Answer:

Given:
Length of the guitar wire (L₁) = 30.0 cm = 0.30 m
Frequency, when no finger is placed on it, (f₁) =196 Hz
And (f₂) =220 Hz, (f₃) = 247 Hz, (f₄) = 262 Hz and (f₅) = 294 Hz
The velocity is constant for a medium.
We have:
$f \propto (\frac{1}{L})$
$\Rightarrow \frac{f_{1}}{f_{2}} = \frac{L_{2}}{L_{1}} \Rightarrow \frac{196}{220} = \frac{L_{2}}{0.3} \Rightarrow L_{2} = \frac{196 \times 0.3}{220} = 0.267 m \Rightarrow L_{2} = 26.7 cm$

Again, $f_{3} = 247 Hz$
$\Rightarrow \frac{f_{3}}{f_{1}} = \frac{L_{1}}{L_{3}} \Rightarrow \frac{247}{196} = \frac{0.3}{L_{3}} \Rightarrow L_{3} = 196 \times \frac{0.3}{247} = 0.238 m \Rightarrow L_{3} = 23.8 cm Similarly, L_{4} = 196 \times \frac{0.3}{262} = 0.224 m \Rightarrow L_{4} = 22.4 cm And, L_{5} = 20 cm$

Question 45:

Given:
Length of the guitar wire (L₁) = 30.0 cm = 0.30 m
Frequency, when no finger is placed on it, (f₁) =196 Hz
And (f₂) =220 Hz, (f₃) = 247 Hz, (f₄) = 262 Hz and (f₅) = 294 Hz
The velocity is constant for a medium.
We have:
$f \propto (\frac{1}{L})$
$\Rightarrow \frac{f_{1}}{f_{2}} = \frac{L_{2}}{L_{1}} \Rightarrow \frac{196}{220} = \frac{L_{2}}{0.3} \Rightarrow L_{2} = \frac{196 \times 0.3}{220} = 0.267 m \Rightarrow L_{2} = 26.7 cm$

Again, $f_{3} = 247 Hz$
$\Rightarrow \frac{f_{3}}{f_{1}} = \frac{L_{1}}{L_{3}} \Rightarrow \frac{247}{196} = \frac{0.3}{L_{3}} \Rightarrow L_{3} = 196 \times \frac{0.3}{247} = 0.238 m \Rightarrow L_{3} = 23.8 cm Similarly, L_{4} = 196 \times \frac{0.3}{262} = 0.224 m \Rightarrow L_{4} = 22.4 cm And, L_{5} = 20 cm$

Answer:

Given:
Fundamental frequency (f₀) of the steel wire = 200 Hz
Let the highest harmonic audible to the person be n.

Frequency of the highest harmonic, f' = 14000 Hz
∴ f'=nf₀ ...(1)
$\frac{f'}{f_{0}} = \frac{14000}{200} \Rightarrow \frac{n f_{0}}{f_{0}} = 70 \Rightarrow n = 70$
Thus, the highest harmonic audible to man is the 70^th harmonic.

Question 46:

Given:
Fundamental frequency (f₀) of the steel wire = 200 Hz
Let the highest harmonic audible to the person be n.

Frequency of the highest harmonic, f' = 14000 Hz
∴ f'=nf₀ ...(1)
$\frac{f'}{f_{0}} = \frac{14000}{200} \Rightarrow \frac{n f_{0}}{f_{0}} = 70 \Rightarrow n = 70$
Thus, the highest harmonic audible to man is the 70^th harmonic.

Answer:

Given:
Let the three resonant frequencies of a string be
$f_{1} = 90 Hz f_{2} = 150 Hz f_{3} = 210 Hz$

(a) So, the highest possible fundamental frequency of the string is $f = 30 Hz$ , because f₁, f₂ and f₃ are the integral multiples of 30 Hz.

(b) So, these frequencies can be written as follows:
$f_{1} = 3 f f_{2} = 5 f f_{3} = 7 f$

Hence, f₁, f₂, and f₃ are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are f, 2f, 3f, 4f, 5f ...
∴ 3f = 2nd overtone and 3rd harmonic
5f = 4th overtone and 5th harmonic
7th= 6th overtone and 7th harmonic

(d) Length of the string (L) = 80 cm = 0.8 m
Let the speed of the wave be v.
So, the frequency of the third harmonic is given by:
$f_{1} = (\frac{3}{2 \times L}) v \Rightarrow 90 = \{\frac{3}{(2 \times 80)}\} \times v \Rightarrow v = \frac{(90 \times 2 \times 80)}{3} = 30 \times 2 \times 80 = 4800 cm / s \Rightarrow v = 48 m / s$

Question 47:

Given:
Let the three resonant frequencies of a string be
$f_{1} = 90 Hz f_{2} = 150 Hz f_{3} = 210 Hz$

(a) So, the highest possible fundamental frequency of the string is $f = 30 Hz$ , because f₁, f₂ and f₃ are the integral multiples of 30 Hz.

(b) So, these frequencies can be written as follows:
$f_{1} = 3 f f_{2} = 5 f f_{3} = 7 f$

Hence, f₁, f₂, and f₃ are the third harmonic, the fifth harmonic and the seventh harmonic, respectively.
(c) The frequencies in the string are f, 2f, 3f, 4f, 5f ...
∴ 3f = 2nd overtone and 3rd harmonic
5f = 4th overtone and 5th harmonic
7th= 6th overtone and 7th harmonic

(d) Length of the string (L) = 80 cm = 0.8 m
Let the speed of the wave be v.
So, the frequency of the third harmonic is given by:
$f_{1} = (\frac{3}{2 \times L}) v \Rightarrow 90 = \{\frac{3}{(2 \times 80)}\} \times v \Rightarrow v = \frac{(90 \times 2 \times 80)}{3} = 30 \times 2 \times 80 = 4800 cm / s \Rightarrow v = 48 m / s$

Answer:

Given:
The tensions in the two wires are in the ratio of 2:1.
$\Rightarrow \frac{T_{1}}{T_{2}} = 2$
Ratio of the radii is 3:1.
$\Rightarrow \frac{r_{1}}{r_{2}} = 3 = \frac{D_{1}}{D_{2}}$
Density in the ratios of 1:2.
$\Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{1}{2}$
Let the length of the wire be L.
$Frequency, f = \frac{1}{L D} \sqrt{\frac{T}{π ρ}}$
$\Rightarrow f_{1} = \frac{1}{L D_{1}} \sqrt{\frac{T_{1}}{π ρ_{1}}} \Rightarrow f_{2} = \frac{1}{L D_{2}} \sqrt{\frac{T_{2}}{π ρ_{2}}}$
$∴ \frac{f_{1}}{f_{2}} = \frac{L D_{2}}{L D_{1}} \sqrt{\frac{T_{1}}{T_{2}}} \sqrt{\frac{π ρ_{2}}{π ρ_{1}}} \Rightarrow \frac{f_{1}}{f_{2}} = \frac{1}{3} \sqrt{\frac{2}{1} \times \frac{2}{1}} \Rightarrow f_{1} : f_{2} = 2 : 3$

Question 48:

Given:
The tensions in the two wires are in the ratio of 2:1.
$\Rightarrow \frac{T_{1}}{T_{2}} = 2$
Ratio of the radii is 3:1.
$\Rightarrow \frac{r_{1}}{r_{2}} = 3 = \frac{D_{1}}{D_{2}}$
Density in the ratios of 1:2.
$\Rightarrow \frac{ρ_{1}}{ρ_{2}} = \frac{1}{2}$
Let the length of the wire be L.
$Frequency, f = \frac{1}{L D} \sqrt{\frac{T}{π ρ}}$
$\Rightarrow f_{1} = \frac{1}{L D_{1}} \sqrt{\frac{T_{1}}{π ρ_{1}}} \Rightarrow f_{2} = \frac{1}{L D_{2}} \sqrt{\frac{T_{2}}{π ρ_{2}}}$
$∴ \frac{f_{1}}{f_{2}} = \frac{L D_{2}}{L D_{1}} \sqrt{\frac{T_{1}}{T_{2}}} \sqrt{\frac{π ρ_{2}}{π ρ_{1}}} \Rightarrow \frac{f_{1}}{f_{2}} = \frac{1}{3} \sqrt{\frac{2}{1} \times \frac{2}{1}} \Rightarrow f_{1} : f_{2} = 2 : 3$

Answer:

Given:
Length of the rod (L) = 40 cm = 0.40 m
Mass of the rod (m) = 1.2 kg
Let the mass of 4.8 kg be placed at x distance from the left.
As per the question, frequency on the left side = f₀
Frequency on the right side = 2f₀
Let tension be T₁ and T₂on the left and the right side, respectively.

$∴ \frac{1}{2 L} \sqrt{\frac{T_{1}}{m}} = \frac{2}{2 L} \sqrt{\frac{T_{2}}{m}} \Rightarrow \sqrt{\frac{T_{1}}{T_{2}}} = 2 \Rightarrow \frac{T_{1}}{T_{2}} = 4 . . . (1)$
From the free body diagram:

$T_{1} + T_{2} = 48 + 12 = 60 N \Rightarrow 4 T_{2} + T_{2} = 5 T_{2} = 60 N [using equation (1)] ∴ T_{2} = 12 N and T_{1} = 48 N$

Now, taking moment about point A:

$T_{2} \times (0.4) = 48 x + 12 \times 0.2 \Rightarrow 4.8 = 48 x - 2.4 \Rightarrow 4.8 x = 2.4 \Rightarrow x = \frac{2.4}{4.8} = \frac{1}{20} m = 5 cm$
Therefore, the mass should be placed at a distance of 5 cm from the left end.

Question 49:

Given:
Length of the rod (L) = 40 cm = 0.40 m
Mass of the rod (m) = 1.2 kg
Let the mass of 4.8 kg be placed at x distance from the left.
As per the question, frequency on the left side = f₀
Frequency on the right side = 2f₀
Let tension be T₁ and T₂on the left and the right side, respectively.

$∴ \frac{1}{2 L} \sqrt{\frac{T_{1}}{m}} = \frac{2}{2 L} \sqrt{\frac{T_{2}}{m}} \Rightarrow \sqrt{\frac{T_{1}}{T_{2}}} = 2 \Rightarrow \frac{T_{1}}{T_{2}} = 4 . . . (1)$
From the free body diagram:

$T_{1} + T_{2} = 48 + 12 = 60 N \Rightarrow 4 T_{2} + T_{2} = 5 T_{2} = 60 N [using equation (1)] ∴ T_{2} = 12 N and T_{1} = 48 N$

Now, taking moment about point A:

$T_{2} \times (0.4) = 48 x + 12 \times 0.2 \Rightarrow 4.8 = 48 x - 2.4 \Rightarrow 4.8 x = 2.4 \Rightarrow x = \frac{2.4}{4.8} = \frac{1}{20} m = 5 cm$
Therefore, the mass should be placed at a distance of 5 cm from the left end.

Answer:

Given:
Length of the aluminium wire (L_a)= 60 cm = 0.60 m
Length of the steel wire (L_s)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (A_a) = 1.0 mm²
Area of cross-section of the steel wire (A_s) = 3.0 mm²
Density of aluminium (ρ_a) = 2⋅6 g cm⁻³
Density of steel (ρ_s) = 7⋅8 g cm⁻³

$Mass per unit length of the steel, m_{s} = ρ_{s} \times A_{s} = 7.8 \times 10^{- 2} gm / cm = 7.8 \times 10^{- 3} kg / m Mass per unit length of the aluminium, m_{A} = ρ_{A} A_{A} = 2.6 \times 10^{- 2} \times 3 gm / cm = 7.8 \times 10^{- 2} gm / cm = 7.8 \times 10^{- 3} kg / m$

A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.
$\Rightarrow v = (\frac{T}{m}) = {\sqrt{\{\frac{40}{7.8 \times 10^{- 3}}\}}}^{-} = \sqrt{(\frac{4 \times 10^{4}}{7.8})} \Rightarrow v = 71.6 m / s$
For minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm
$\Rightarrow Wavelength, λ = 2 \times 20 = 40 cm Or λ = 0.4 m ∴ Frequency, f = \frac{v}{λ} = \frac{71.6}{0.4} = 180 Hz$

Question 50:

Given:
Length of the aluminium wire (L_a)= 60 cm = 0.60 m
Length of the steel wire (L_s)= 80 cm = 0.80 m
Tension produced (T) = 40 N
Area of cross-section of the aluminium wire (A_a) = 1.0 mm²
Area of cross-section of the steel wire (A_s) = 3.0 mm²
Density of aluminium (ρ_a) = 2⋅6 g cm⁻³
Density of steel (ρ_s) = 7⋅8 g cm⁻³

$Mass per unit length of the steel, m_{s} = ρ_{s} \times A_{s} = 7.8 \times 10^{- 2} gm / cm = 7.8 \times 10^{- 3} kg / m Mass per unit length of the aluminium, m_{A} = ρ_{A} A_{A} = 2.6 \times 10^{- 2} \times 3 gm / cm = 7.8 \times 10^{- 2} gm / cm = 7.8 \times 10^{- 3} kg / m$

A node is always placed at the joint. Since aluminium and steel rod has same mass per unit length, the velocity of wave (v) in both of them is same.
Let v be the velocity of wave.
$\Rightarrow v = (\frac{T}{m}) = {\sqrt{\{\frac{40}{7.8 \times 10^{- 3}}\}}}^{-} = \sqrt{(\frac{4 \times 10^{4}}{7.8})} \Rightarrow v = 71.6 m / s$
For minimum frequency, there would be maximum wavelength.
For maximum wavelength, minimum number of loops are to be produced.
∴ Maximum distance of a loop = 20 cm
$\Rightarrow Wavelength, λ = 2 \times 20 = 40 cm Or λ = 0.4 m ∴ Frequency, f = \frac{v}{λ} = \frac{71.6}{0.4} = 180 Hz$

Answer:

Given:
Length of the string = L
Velocity of wave is given as:
$V = \sqrt{\frac{T}{m}}$
In fundamental mode, frequency = ν
$\Rightarrow ν = \frac{1}{2 L} \sqrt{\frac{T}{m}}$
(a) $Wavelength, λ = \frac{V elocity}{F requency}$
$\Rightarrow λ = \frac{\sqrt{\frac{T}{m}}}{\frac{1}{2 L} \sqrt{\frac{T}{m}}} = 2 L Wave number, K = \frac{2 π}{λ} = \frac{2 π}{2 L} = \frac{π}{L}$

(b) Equation of the stationary wave:
$y = A \cos (\frac{2 π x}{λ}) \sin (\frac{2 π V t}{λ}) = A \cos (\frac{2 π x}{2 L}) \sin (\frac{2 π V t}{2 L}) [∵ ν = (\frac{V}{2 L})] = A \cos (\frac{π x}{L}) \sin (2 π ν t)$

Question 51:

Given:
Length of the string = L
Velocity of wave is given as:
$V = \sqrt{\frac{T}{m}}$
In fundamental mode, frequency = ν
$\Rightarrow ν = \frac{1}{2 L} \sqrt{\frac{T}{m}}$
(a) $Wavelength, λ = \frac{V elocity}{F requency}$
$\Rightarrow λ = \frac{\sqrt{\frac{T}{m}}}{\frac{1}{2 L} \sqrt{\frac{T}{m}}} = 2 L Wave number, K = \frac{2 π}{λ} = \frac{2 π}{2 L} = \frac{π}{L}$

(b) Equation of the stationary wave:
$y = A \cos (\frac{2 π x}{λ}) \sin (\frac{2 π V t}{λ}) = A \cos (\frac{2 π x}{2 L}) \sin (\frac{2 π V t}{2 L}) [∵ ν = (\frac{V}{2 L})] = A \cos (\frac{π x}{L}) \sin (2 π ν t)$

Answer:

Given:
Length of the string (L) = 2.0 m
Wave speed on the string in its first overtone (v) = 200 m/s
Amplitude (A) = 0.5 cm

(a) Wavelength and frequency of the string when it is vibrating in its 1^st overtone (n = 2):
$L = \frac{n λ}{2}$
$\Rightarrow λ = L = 2 m \Rightarrow f = \frac{ν}{λ} = \frac{200}{2} = 100 Hz$
(b) The stationary wave equation is given by:

$y = 2 A \cos \frac{2 π x}{λ} \sin \frac{2 π v t}{λ} = 0.5 \cos \frac{2 π x}{2} \sin \frac{2 π \times 200 t}{2} = (0.5 cm) \cos [(π m^{- 1}) x] \sin [(200 π s^{- 1}) t]$

Question 52:

Given:
Length of the string (L) = 2.0 m
Wave speed on the string in its first overtone (v) = 200 m/s
Amplitude (A) = 0.5 cm

(a) Wavelength and frequency of the string when it is vibrating in its 1^st overtone (n = 2):
$L = \frac{n λ}{2}$
$\Rightarrow λ = L = 2 m \Rightarrow f = \frac{ν}{λ} = \frac{200}{2} = 100 Hz$
(b) The stationary wave equation is given by:

$y = 2 A \cos \frac{2 π x}{λ} \sin \frac{2 π v t}{λ} = 0.5 \cos \frac{2 π x}{2} \sin \frac{2 π \times 200 t}{2} = (0.5 cm) \cos [(π m^{- 1}) x] \sin [(200 π s^{- 1}) t]$

Answer:

Given:
The stationary wave equation of a string vibrating in its third harmonic is given by
y = (0.4 cm) sin [(0.314 cm⁻¹) x]cos [(.600 πs⁻¹) t]
By comparing with standard equation,
$y = A \sin (k x) \cos (w t)$

(a) From the above equation, we can infer the following:
$ω = 600 π$
$\Rightarrow 2 π f = 600 π \Rightarrow f = 300 Hz$
Wavelength, $λ = \frac{2 π}{0.314} = \frac{(2 \times 3.14)}{0.314}$
$\Rightarrow λ = 20 cm$

(b) Therefore, the nodes are located at 0cm, 10 cm, 20 cm, 30 cm.

(c) Length of the string, l = $n \frac{λ}{2}$
$\Rightarrow l = \frac{3 λ}{2} = \frac{3 \times 20}{2} = 30 cm$

(d) $y = 0.4 \sin (0.314 x) \cos (600 π t)$
$= 0.4 \sin \{(\frac{π}{10}) x\} \cos (600 π t)$

$λ$ and $ν$ are the wavelength and velocity of the waves that interfere to give this vibration.

$λ = 20 cm ν = \frac{ω}{k} = \frac{600 π}{(\frac{π}{10})} = 6000 cm / s \Rightarrow ν = 60 m / s$

Question 53:

Given:
The stationary wave equation of a string vibrating in its third harmonic is given by
y = (0.4 cm) sin [(0.314 cm⁻¹) x]cos [(.600 πs⁻¹) t]
By comparing with standard equation,
$y = A \sin (k x) \cos (w t)$

(a) From the above equation, we can infer the following:
$ω = 600 π$
$\Rightarrow 2 π f = 600 π \Rightarrow f = 300 Hz$
Wavelength, $λ = \frac{2 π}{0.314} = \frac{(2 \times 3.14)}{0.314}$
$\Rightarrow λ = 20 cm$

(b) Therefore, the nodes are located at 0cm, 10 cm, 20 cm, 30 cm.

(c) Length of the string, l = $n \frac{λ}{2}$
$\Rightarrow l = \frac{3 λ}{2} = \frac{3 \times 20}{2} = 30 cm$

(d) $y = 0.4 \sin (0.314 x) \cos (600 π t)$
$= 0.4 \sin \{(\frac{π}{10}) x\} \cos (600 π t)$

$λ$ and $ν$ are the wavelength and velocity of the waves that interfere to give this vibration.

$λ = 20 cm ν = \frac{ω}{k} = \frac{600 π}{(\frac{π}{10})} = 6000 cm / s \Rightarrow ν = 60 m / s$

Answer:

Given:
Equation of the standing wave:
$y = (0.4 cm) \sin [(0.314 {cm}^{- 1}) x] \cos [(600 π s^{- 1}) t] \Rightarrow k = 0.314 = \frac{π}{10} Also, k = \frac{2 π}{λ} \Rightarrow λ = 20 cm$

We know:
$L = \frac{n λ}{2}$
For the smallest length, putting n = 1:
$\Rightarrow L = \frac{λ}{2} = \frac{20 cm}{2} = 10 cm$
Therefore, the required length of the string is 10 cm.

Question 54:

Given:
Equation of the standing wave:
$y = (0.4 cm) \sin [(0.314 {cm}^{- 1}) x] \cos [(600 π s^{- 1}) t] \Rightarrow k = 0.314 = \frac{π}{10} Also, k = \frac{2 π}{λ} \Rightarrow λ = 20 cm$

We know:
$L = \frac{n λ}{2}$
For the smallest length, putting n = 1:
$\Rightarrow L = \frac{λ}{2} = \frac{20 cm}{2} = 10 cm$
Therefore, the required length of the string is 10 cm.

Answer:

Given:
Length of the wire (L) = 40 cm = 0.40 m
Mass of the wire = 3.2 g = 0.003 kg
Distance between the two fixed supports of the wire = 40.05 cm
Fundamental mode frequency = 220 Hz
Therefore, linear mass density of the wire (m) is given by:
$m = \frac{0.0032}{0.4} = 8 \times 10^{- 3} kg / m Change in length, ∆ L = 40.05 - 40 = 0.05 cm = 0.05 \times 10^{- 2} m Strain = \frac{∆ L}{L} = \frac{0.05 \times 10^{- 2}}{0.4} = 0.125 \times 10^{- 2} f_{0} = \frac{1}{2 L} \sqrt{\frac{T}{m}} = \frac{1}{2 \times (0.4005)} \sqrt{\frac{T}{8 \times 10^{- 3}}}$

$\Rightarrow 220 \times 220 = [\frac{1}{{(0.801)}^{2}}] \times T \times (\frac{10^{3}}{8}) \Rightarrow T \times 1000 = 220 \times 220 \times 0.641 \times 0.8 \Rightarrow T = 248.19 N Stress = \frac{Tension}{Area} = \frac{248.19}{1 {mm}^{2}} = \frac{248.19}{10^{- 6}} \Rightarrow Stress = 248.19 \times 10^{6} Young' s modulus, Y = \frac{stress}{strain} = \frac{248.19 \times 10^{6}}{0.125 \times 10^{- 2}} \Rightarrow Y = 19852 \times 10^{8} = 1.985 \times 10^{11} N / m^{2}$
Hence, the required Young's modulus of the wire is $1.985 \times 10^{11} N / m^{2}$ .

Question 55:

Given:
Length of the wire (L) = 40 cm = 0.40 m
Mass of the wire = 3.2 g = 0.003 kg
Distance between the two fixed supports of the wire = 40.05 cm
Fundamental mode frequency = 220 Hz
Therefore, linear mass density of the wire (m) is given by:
$m = \frac{0.0032}{0.4} = 8 \times 10^{- 3} kg / m Change in length, ∆ L = 40.05 - 40 = 0.05 cm = 0.05 \times 10^{- 2} m Strain = \frac{∆ L}{L} = \frac{0.05 \times 10^{- 2}}{0.4} = 0.125 \times 10^{- 2} f_{0} = \frac{1}{2 L} \sqrt{\frac{T}{m}} = \frac{1}{2 \times (0.4005)} \sqrt{\frac{T}{8 \times 10^{- 3}}}$

$\Rightarrow 220 \times 220 = [\frac{1}{{(0.801)}^{2}}] \times T \times (\frac{10^{3}}{8}) \Rightarrow T \times 1000 = 220 \times 220 \times 0.641 \times 0.8 \Rightarrow T = 248.19 N Stress = \frac{Tension}{Area} = \frac{248.19}{1 {mm}^{2}} = \frac{248.19}{10^{- 6}} \Rightarrow Stress = 248.19 \times 10^{6} Young' s modulus, Y = \frac{stress}{strain} = \frac{248.19 \times 10^{6}}{0.125 \times 10^{- 2}} \Rightarrow Y = 19852 \times 10^{8} = 1.985 \times 10^{11} N / m^{2}$
Hence, the required Young's modulus of the wire is $1.985 \times 10^{11} N / m^{2}$ .

Answer:

Density of the block = ρ
Volume of block = V
∴ Weight of the block is, W = ρVg
∴ Tension in the string, T = W

The tuning fork resonates with different frequencies in the two cases.
Let the tenth harmonic be f₁₀.

$f_{10} = \frac{10}{2 L} \sqrt{\frac{T}{m}} = \frac{10}{2 L} \sqrt{\frac{ρ V g}{m}}$
Here, m is the mass per unit length of the string and L is the length of the string.

When the block is immersed in water (density = ρ ), let the eleventh harmonic be f₁₁.

$f_{11} = \frac{11}{2 L} \sqrt{\frac{T'}{m}} = \frac{11}{2 L} \sqrt{\frac{(ρ - ρ_{w}) V g}{m}}$
The frequency (f) of the tuning fork is same.
$∴ f_{10} = f_{11} \Rightarrow \frac{10}{2 L} \sqrt{\frac{ρ V g}{m}} = \frac{11}{2 L} \sqrt{\frac{(ρ - ρ_{ω}) V g}{m}} \Rightarrow \frac{100}{121} = \frac{ρ - 1}{ρ} (because ρ_{ω} = 1 gm / cc) \Rightarrow 100 ρ = 121 ρ - 121 \Rightarrow ρ = \frac{121}{21} = 5.8 gm / cc = 5.8 \times 10^{3} kg / m^{3}$
Therefore, the required density is $5.8 \times 10^{3} kg / m^{3}$ .

Question 56:

Density of the block = ρ
Volume of block = V
∴ Weight of the block is, W = ρVg
∴ Tension in the string, T = W

The tuning fork resonates with different frequencies in the two cases.
Let the tenth harmonic be f₁₀.

$f_{10} = \frac{10}{2 L} \sqrt{\frac{T}{m}} = \frac{10}{2 L} \sqrt{\frac{ρ V g}{m}}$
Here, m is the mass per unit length of the string and L is the length of the string.

When the block is immersed in water (density = ρ ), let the eleventh harmonic be f₁₁.

$f_{11} = \frac{11}{2 L} \sqrt{\frac{T'}{m}} = \frac{11}{2 L} \sqrt{\frac{(ρ - ρ_{w}) V g}{m}}$
The frequency (f) of the tuning fork is same.
$∴ f_{10} = f_{11} \Rightarrow \frac{10}{2 L} \sqrt{\frac{ρ V g}{m}} = \frac{11}{2 L} \sqrt{\frac{(ρ - ρ_{ω}) V g}{m}} \Rightarrow \frac{100}{121} = \frac{ρ - 1}{ρ} (because ρ_{ω} = 1 gm / cc) \Rightarrow 100 ρ = 121 ρ - 121 \Rightarrow ρ = \frac{121}{21} = 5.8 gm / cc = 5.8 \times 10^{3} kg / m^{3}$
Therefore, the required density is $5.8 \times 10^{3} kg / m^{3}$ .

Answer:

Given:
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density, m $= \frac{0.08}{2} = 0.04 kg / m$
$Tension, T = 256 N Wave velocity, v = \sqrt{\frac{T}{m}} \Rightarrow v = \sqrt{(\frac{256}{0.04})} = \frac{160}{2} \Rightarrow v = 80 m / s$

For fundamental frequency:
$L = \frac{λ}{4} \Rightarrow λ = 4 L = 4 \times 2 = 8 m \Rightarrow f = \frac{v}{λ} = \frac{80}{8} = 10 Hz$

(a) The frequency overtones are given below:
$1 st overtone = 3 f = 30 Hz 2 nd overtone = 5 f = 50 Hz$

(b) $λ = 4 l = 4 \times 2 = 8 m$
$∴ λ_{1} = \frac{v}{f_{1}} = \frac{80}{30} = 2.67 m λ_{2} = \frac{v}{f_{2}} = \frac{80}{50} = 1.6 m$

Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.

Question 57:

Given:
Length of the long rope (L) = 2.00 m
Mass of the rope = 80 g = 0.08 kg
Tension (T) = 256 N
Linear mass density, m $= \frac{0.08}{2} = 0.04 kg / m$
$Tension, T = 256 N Wave velocity, v = \sqrt{\frac{T}{m}} \Rightarrow v = \sqrt{(\frac{256}{0.04})} = \frac{160}{2} \Rightarrow v = 80 m / s$

For fundamental frequency:
$L = \frac{λ}{4} \Rightarrow λ = 4 L = 4 \times 2 = 8 m \Rightarrow f = \frac{v}{λ} = \frac{80}{8} = 10 Hz$

(a) The frequency overtones are given below:
$1 st overtone = 3 f = 30 Hz 2 nd overtone = 5 f = 50 Hz$

(b) $λ = 4 l = 4 \times 2 = 8 m$
$∴ λ_{1} = \frac{v}{f_{1}} = \frac{80}{30} = 2.67 m λ_{2} = \frac{v}{f_{2}} = \frac{80}{50} = 1.6 m$

Hence, the wavelengths are 8 m, 2.67 m and 1.6 m, respectively.

Answer:

Let T be the tension in the string and m be the mass per unit length of the heavy string.
In the first part of the question, the heavy string is fixed at only one end.
So, the lowest frequency is given by:
$f_{0} = \frac{1}{4 L} \sqrt{\frac{T}{m}} . . . (1)$
When the movable support is pushed by 10 cm to the right, the joint is placed on the pulley and the heavy string becomes fixed at both the ends (keeping T and m same).
Now, the lowest frequency is given by:
$f_{0}' = \frac{1}{2 L} \sqrt{\frac{T}{m}} . . . (2)$
Dividing equation (2) by equation (1), we get:
$f_{0}' = 2 f_{0} = 240 Hz$

Wave Motion and Wave on a String

Class 11th Concepts Of Physics Part 1 HC Verma Solution

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Class 11^th Concepts Of Physics Part 1 HC Verma Solution