Waves

Given:

Mass of the string, M = 2.50 Kg

Tension in the string, T = 200 N

Length of the string, L = 20 m

Mass per unit length, m = M/L

⇒ m = 2.50 Kg/ 20m

⇒ m = 0.125 Kgm^-1

The velocity, V of the transverse wave in the string is given by as,

V = (T/m)^-2 …(1)

Where,

T = tension in the string

m = mass per unit length

By putting the given data in equation (1), we get,

⇒ v = 40 ms^-1

Time, t taken for the wave to get to the other end can be given by,

t = L/V …(2)

Where,

L = length of the string

V = velocity of wave

Using equation (2),

t = 20m/ 40 ms^-1

t = 0.5 s

Given:

Height of the tower, h = 300 m

Initial velocity of the stone, u = 0 ms^-1

Acceleration due to gravity, g = 9.8 ms^-1

Speed of the sound in air = 340 ms^-1

The time taken by the stone to hit the pond can be found by the second equation of motion,

s = ut + 1/2 gt² …(1)

Where,

s = distance, in this case height

u = initial velocity

g = acceleration due to gravity

t = time

Putting the values in equation (1), we get

s = ut + 1/2 gt²

t = (2s/g)^1/2

t = [2× (300m/9.8 ms^-2)]^1/2

t = 7.82 s

Time taken by the sound of splash to the top of the tower can also be given by,

t₁ = h/v …(2)

Where,

h = height of the tower

v = velocity of sound (340 m/s)

t₁ = 300m/340ms^-1

t₁ = 0.88 s

The total time after which splash is heard is equal to sum of the time taken by the stone to reach the water and time taken by sound to reach the tower.

i.e. T = t + t₁

T = 7.82s + 0.88 s

T = 8.7 s

Given:

Mass of the steel wire, M = 2.10 Kg

Length of the string, L = 12 m

Mass per unit length, m = M/L

⇒ m = 2.10 Kg/ 12m

⇒ m = 0.175 Kgm^-1

Velocity of the transverse wave, V = 343ms^-1

The Tension, T in transverse wave in the string is given by,

T = V²m …(1)

Where,

T = tension in the string

m = mass per unit length

V = velocity of wave

By putting the given data in equation (1), we get,

T = (343ms^-1)²× 0.175Kgm^-1

⇒ T = 20588.5 N

(a) Given:

…(1)

Where,

γ = adiabatic coefficient / index

P = pressure

ρ = Density of medium

ρ = Mass (M) /Volume (V)

By putting the value of density, in equation (1), it can be reduced to,

…(2)

Ideal gas equation is written as,

PV = RT …(3) (for n = 1)

Since, R is a constant, For constant Temperature, PV is constant.

From equation (2), we see that γ and M are also constant,

So it can be concluded that, velocity of sound is constant at constant temperature, it is not affected by change in pressure.

(b)

Ideal gas equation is written as,

PV = RT (for n = 1)

P = RT/V

The equation of velocity can be rewritten as,

…(5)

We know that V×ρ (i.e. mass), γ and R are constants.

We can write the equation (5) as,

Hence the speed of sound in a medium is proportional to the square root of the temperature of the gaseous medium.

(c) Let v_m, ρ_m and v_d, ρ_d be the speed of the sound and density in the moist air and dry air respectively.

From the equation,

The velocity in moist air and dry air can be written as

…(6)

…(7)

By dividing equation (6) by equation (7), we get,

…(8)

The presence of moisture in air reduces the density of air,

i.e. ρ_d> ρ_m

so, v_m>v_d.

Hence, the speed of sound in moist air is greater than speed in dry air. Thus in gaseous medium, the speed of sound increases with humidity.

The converse is not true.

A wave function for a travelling wave is used to represent a travelling wave in the terms of x and t, wave function should also have finite value. Of the given alternatives, none can represent a wave function for a travelling wave.

Given:

(a) Frequency of the ultrasonic sound, v = 1000kHz = 10⁶Hz

Speed of sound in air, v_a = 340ms^-1

The wavelength (λ_r) of the reflected sound is given by the relation:

λ_r = v_a/v

λ_r = 340ms^-1/ 10⁶s^-1

λ_r = 3.4 × 10^-4 m

(b) Frequency of the ultrasonic sound, v = 1000 KHz

Speed of the sound in water, v_w = 1486 ms^-1

The wavelength (λ_r) of the reflected sound is given by the relation:

λ_r = v_w/v

λ_r = 1486ms^-1/ 10⁶s^-1

λ_r = 1.488 × 10^-3 m

Given:

Speed of sound in the tissue, v_s = 1.7 Km/s or 1.7× 10³ ms^-1

Operating frequency of the scanner, v = 4.2 MHz or 4.2× 10⁶Hz

The wavelength of the sound in the tissue is given as,

λ = v_s/v

λ = 1700ms^-1/ 4.2× 10⁶s^-1

λ = 4.1× 10^-4m

Given:

Wave function, y = 3.0 sin (36 t + 0.018 x + π/4) …(1)

(a) The equation of a travelling wave is given by,

y(x,t) = a× sin(wt + kx + θ) …(2)

The given equation (1) is in the form of equation (2),

By comparing equation (1) and equation (2), we get that,

k = 0.018 m^-1

w = 36 rad/sec

We know that,

v = w/2π …(3)

Where,

v = frequency

w = angular velocity

λ = 2π/k …(4)

velocity, V = vλ …(5)

By combing equation (3) and (4) we get,

v = w/k

v = 36rads^-1/0.018m^-1

v = 20ms^-1

The velocity of the wave is 20 m/s.

(b) Amplitude a, of the given wave can be found by comparing equation 1 and 2.

a = 3cm = 0.03 m

Frequency, v = w/2π

v = 35rad s^-1/(2× 3.14)

v = 573 Hz

(c) Initial phase angle can be found by comparing equation (1) and (2),

θ = π/4

(d) The distance between consecutive crests or troughs is the wavelength of the wave, it is given by,

λ = 2π/k

λ = 2π/0.018m^-1

λ = 3.49 m

Given:

Wave function, y = 3.0 sin (36 t + 0.018 x + π/4) …(1)

For x = 0, equation reduces to,

y = 3.0 sin (36 t + π/4) …(2)

w = 2π/T

T = 2π rad/36rad s^-1

T = π/18

In the same fashion we can also find the graph for x = 2 and x = 4, for these cases the phase will change.

Given:

Equation of a travelling wave is given by

y(x, t) = 2.0 cos 2π (10t – 0.0080 x + 0.35)

which can be written as,

y(x, t) = 2.0 cos(20 πt – 0.016πx + 0.7 π )

Where,

Propagation constant, k = 0.016 π

Amplitude, a = 2 cm

Angular frequency, w = 20 π rad s^-1

Phase difference is given by the relation,

Φ = kx = 2 π/λ …(1)

a) For x = 4m = 400 cm

Φ = 0.016 π× 400cm = 6.4 π rad

b) For x = 0.5m = 50 cm

Φ = 0.016 π× 50cm = 0.8 π rad

c) For x = λ/2

Φ = 2 π/λ × λ/2 = π rad

d) For x = 3λ/4

Φ = 2 π/λ × 3λ/4 = 3π/2 rad

Given:

…(1)

(a) The general equation of an standing wave is given by,

y(x, t) = 2 sin kx × cos wt …(2)

The equation 1, is similar to equation 2, hence the given equation represents a standing wave.

(b) A wave travelling along the positive x-direction is given by,

y₁ = a sin(wt-kx)

For a wave travelling in negative x direction is given by,

y₂ = a sin (wt + kx)

The superposition of these two waves can be written as,

y = y₁ + y₂

y = a sin (wt-kx) – a sin(wt + kx)

The terms can be expanded using the formula of sin(a + b) and the result is,

y = 2 a sin(kx) cos(wt)

y = 2a sin (2πx/λ) cos(2πvt) …(3)

By comparing equation (1) and (3) we have,

2 π/λ = 2 π/3

⇒ wavelength, λ = 3 m

It is given that,

120 π = 2 πv

Frequency, v = 60Hz

Wave speed, V is given by,

V = v× λ

V = 60 s^-1× 3m

V = 180 ms^-1

(c) The velocity of a transverse wave in a string is given by,

V = (T/m)^1/2

Where,

V = velocity of wave,

T = tension in string

m = mass per unit length of string

⇒ m = mass of the string / length of the string

⇒ m = 3.02× 10^-2Kg/ 1.5 m

⇒ m = 2× 10^-2 Kgm^-1

Tension in the string, T = V²m

T = 2× 10^-2Kgm^-1× (180ms^-1)²

T = 648 N

(i) For the wave on the string described in the question, we have seen that, L = 1.5 m and λ = 3 m. It is clear that L = λ/2 and for a string clamped at both ends, it is possible only if both the ends behave as nodes and there is one anti-node in between.

a) Yes, all the string particles except the nodes, vibrate with the same frequency, v = 60 Hz.

b) As all string particles lie in one segment, all of them are in same phase.

c) Amplitude varies from particle to particle, at anti-nodes amplitude is 2a i.e. 0.06 m. It gradually falls towards nodes. At the nodes the amplitude is zero.

(ii) Distance from one end, x = 0.375 m

Amplitude at the distance of 0.375 is given by,

A = 0.042 m

(a) It represents a stationary wave.

(b) It doesn’t represent either a travelling wave or a stationery wave.

(c) It is a representation for the travelling wave.

(d) It represents the superposition of two stationery waves.

(a) Given,

Mass of the wire, m = 3.5 × 10^–2 kg

Linear mass density, μ = m/l = 4.0 × 10^-2 kg m^-1

Frequency of vibration, f = 45 Hz

So, Length of the wire, l = m/μ

⇒ l = (3.5 × 10^–2 kg)/(4.0 × 10^-2 kg m^-1)

⇒ l = 0.875 m

The wavelength of the stationary wave (λ) and the length of the wire are related as

λ = 2l/m

where, n = Number of nodes in the wire

For fundamental node, n = 1

∴ λ = 2l

⇒ λ = 2 × 0.875 m

⇒ λ = 1.75 m

The speed of the transverse wave in the string is given as

V = fλ= 45 m s^-1 × 1.75 m

⇒ V = 78.75 m s^-1

(b) The tension produced in the string is given as

T = v²μ = (78.75 m s^-1)² × 4.0 × 10^–2 kg m^-1

⇒ T = 248.06 N

Given,

Frequency of the turning fork, f = 340 Hz

Length of pipe, l = 25.5 cm = 0.255 m

Since the given pipe is attached with a piston at one end, one of its end is closed. So, it will behave as a pipe with one end closed and the other end open. Such a system produces odd harmonics.

The fundamental note in a closed pipe is given by

l = π/4

∴ λ = 4l

⇒ λ = 4 × 0.255 m

⇒ λ = 1.02 m

The speed of the sound is given by

v = fλ

⇒ v = 340 Hz × 1.02 m

⇒ v = 346.8 m s^-1

Given,

Length of the steel rod, L = 100 cm = 1 m

Fundamental frequency of vibration, f = 2.53 kHz = 2.53 × 10³ Hz When the rod is plucked at its mid-point, an antinode (A) is formed at its centre, and nodes (N) are formed at its two ends.

The distance between two successive node is λ/2.

∴ L = λ/2

⇒ λ = 2L

⇒ λ = 2 × 1 m

⇒ λ = 2 m

The speed of sound in steel is given by

v = fλ

⇒ v = 2.53 × 10³ Hz × 2 m

⇒ v = 5.06 × 10³ m s^-1

⇒ v = 5.06 km s^-1

NOTE: A node is a point along a standing wave where the wave has minimum amplitude. The opposite of a node is an anti-node, a point where the amplitude of the standing wave is a maximum.

First (Fundamental) harmonic mode of the pipe will be resonantly excited by a 430 Hz source. The same source will not be in resonance with the pipe if both ends are open.

Explanation:

Given,

Length of the closed pipe, l = 20 cm = 0.2 m

Source frequency = n^th normal mode of frequency, f_n = 430 Hz

Speed of sound, v = 340 m s^-1

In a closed pipe, the nth normal mode of frequency is given by

⇒

⇒ 2n-1 = 1.012

⇒ 2n = 2.012

⇒ n = 1.006 or 1

Hence, the frequency corresponding to the first (fundamental) mode of vibration is resonantly excited by the given source. In a pipe open at both ends, the n^th mode of vibration frequency is given by

f_n = nv/2l

⇒ n = 2lf_n/v

NOTE: Resonance means that a certain frequency of wave hitting some object is "in synch" with the natural vibrating frequency of that object. If the wave is in synch, it reinforces the natural vibration of the object and can cause the amplitude (amount) of vibration of the object to increase greatly.

Given,

Frequency of string A, f_A = 324 Hz

Let the frequency of string B be f_B.

Beat’s frequency, n = 6 Hz

Beat's Frequency is given as:

n=|f_Af_B|

⇒ 6 = 324 f_B

⇒ f_B = 318 Hz or 330 Hz

Frequency decreases with a decrease in the tension in a string because frequency is directly proportional to the square root of tension as f ∝ √T

Hence, the beat frequency cannot be 330 Hz.

So, f_B = 318 Hz

NOTE: A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies.

A node (N) is a point where the amplitude of vibration is the minimum and pressure is the maximum. An antinode (A) is a point where the amplitude of vibration is the maximum and pressure is the minimum. Therefore, a displacement node is also a pressure antinode and vice versa.

Bats emit ultrasonic waves of large frequencies. When these waves are reflected from the obstacles in their path, they give them the idea about the distance, direction, size and nature of the obstacle.

The overtones produced by a sitar and a violin, and the strengths of these overtones, are different. Hence, one can distinguish between the notes produced by a sitar and a violin even if they have the same frequency of vibration.

Solids can support both longitudinal and transverse waves, but only longitudinal waves can propagate in gases. This is because solids have both, the elasticity of volume and elasticity of shape, whereas gases have only the volume elasticity.

A sound pulse is a combination of waves of different wavelength. As waves of different wavelength travel in a dispersive medium with different velocities, the shape of the pulse gets distorted.

(i) (a) Given,

Frequency of the whistle, f = 400 Hz

Speed of the train, v_T= 10 m/s

Speed of sound, v = 340 m/s

The apparent frequency (f') of the whistle as the train approaches the platform is given by

f’ = [v/(v-v_T)]f

⇒ f’ = [340/(340-10)]×400

⇒ f’ = 412.12 Hz

(b) The apparent frequency (f’') of the whistle as the train recedes from the platform is given by

f’’ = [v/(v+v_T)]f

⇒ f’ = [340/(340+10)]×400

⇒ f’ = 388.57 Hz

(ii) The apparent change in the frequency of sound is caused by the relative motions of the source and the observer. These relative motions produce no effect on the speed of sound. Therefore, the speed of sound in air in both the cases remains the same which is equal to 340 m/s.

NOTE: The Doppler effect is the phenomenon due to which there is a change in frequency or wavelength of a wave received by an observer who is moving relative to the wave source.

For the stationary observer:

Frequency of the sound produced by the whistle, f = 400 Hz

Speed of sound, v_s = 340 m/s

Velocity of the wind, v = 10 m/s

As there is no relative motion between the source and the observer, the frequency of the sound heard by the observer will be the same as that produced by the source, i.e., 400 Hz. The wind is blowing toward the observer. Hence, the effective speed of the sound increases by 10 units,

So, effective speed of the sound, v_e = 340 + 10 = 350 m/s

The wavelength (λ) of the sound heard by the observer is given by

λ = v_e/f

⇒ λ = (350 m/s)/(400 Hz)

⇒ λ = 0.875 m

For the running observer:

Velocity of the observer, v_o = 10 m/s

The observer is moving toward the source. As a result of the relative motions of the source and the observer, there is a change in frequency (f') which is given by

f’ = [(v+v_o)/v]×f

⇒ f’ = [(340+10)/340]×400

⇒ f’ = 411.76 Hz

Since the air is still, the effective speed of sound will be the same, i.e., 340 m/s. The source is at rest. Hence, the wavelength of the sound will not change, i.e., λ remains 0.875 m. Hence, the given two situations are not exactly identical.

(a) The given harmonic wave is

y(x,t) = 7.5sin(0.0050x + 12t + π/4)

For x=1cm and t=1s,

y(1,1) = 7.5sin(0.005×1 + 12×1 +)

⇒ y(1,1) = 7.5sin(12.005 + )

⇒ y(1,1) = 7.5sin(12.79)

⇒ y(1,1) = 7.5 × 0.2217

⇒ y(1,1) = 1.663 cm

The velocity of oscillation is calculated as

⇒

⇒

At x=1cm and t=1s,

v(1,1) = 90cos(0.005×1 + 12×1 + )

⇒ v(1,1) = 90cos(12.005 +)

⇒ v(1,1) = 7.5cos(12.79)

⇒ v(1,1) = 7.5 × 0.975

⇒ v(1,1) = 87.75 cm/s

The equation of a propagating wave is given by

y(x,t) = asin(kx + wt + Φ)

where, k =

∴ λ =

and ω = 2πf

∴ f =

Speed, v = fλ =

where, ω = 12rad/s and k = 0.0050 m^-1

∴ v =

⇒ v = 2400 cm/s

Hence, the velocity of the wave oscillation at x = 1 cm and t = 1s is not equal to the velocity of the wave propagation.

(b) Propagation constant is related to wavelength as: k = 2π/λ

∴ λ =

⇒ λ =

⇒ λ = 1256 cm

⇒ λ = 12.56 m

Therefore, all the points at distance nλ (n = 1, 2, .....), i.e. 12.56 m, 25.12 m, … and so on for x = 1 cm, will have the same displacement as the x = 1 cm points at t = 2 s, 5 s, and 11 s.

(a) A short pip be a whistle has neither a definite wavelength nor a definite frequency. However, its speed of propagation is fixed, being equal to speed of sound in air.

(b) No, frequency of the note produced by a whistle is not 1/20 = 0.05 Hz. Rather 0.05 Hz is the frequency of repetition of the short pip of the whistle.

The equation of a travelling wave propagating along the positive y-direction is given by the displacement equation,

y(x,t) = asin(ωt-kx) ………… (i)

Linear mass density, μ = 8.0 × 10^-3 kg m^-1

Frequency of the tuning fork, f = 256 Hz

Amplitude of the wave, a = 5.0 cm = 0.05 m

Mass of the pan, m = 90 kg

Tension in the string, T = mg

⇒ T = 90 kg × 9.8 m s^-2

⇒ T = 882 N

The velocity of the transverse wave (v) is given by the relation

⇒

⇒ v = 332.04 m s^-1

Angular frequency, ω = 2πf

∴ ω = 2 × 3.14 × 256 Hz

⇒ ω = 1608.5 rad/s

Wavelength, λ = v/f

⇒ λ = (332.04 m s^-1)/(256 Hz)

⇒ λ = 1.297 m

Propagation constant, k = 2π/λ

∴ k = (2×3.14)/(1.297m)

⇒ k = 4.84 m^-1

Using these values in equation (i),

y(x,t) = 0.05sin(1608.5t – 4.84x)

Given,

Operating frequency of the SONAR system, f = 40 kHz = 40 × 10³ Hz

Speed of the enemy submarine, v_e = 360 km/h = 100 m/s

Speed of sound in water, v = 1450 m/s

The source is at rest and the observer (enemy submarine in this case) is moving towards it. Hence, the apparent frequency (f') received and reflected by the submarine is given by

f’ = [(v+v_e)/v]f

⇒ f’ = [(1450 m s^-1 + 100 m s^-1)/(1450 m s^-1)]×(40×10³ Hz)

⇒ f’ = 42758.6 Hz

⇒ f’ = 42.759 kHz

The frequency (f’’) received by the enemy submarine is given by

f’’ = [v/(v-v_e)]f’

⇒ f’’ = [(1450 m/s)/(1450 m/s – 100 m/s)]×(42758.6 Hz)

⇒ f’’ = 45925.9 Hz

⇒ f’’ = 45.93 kHz

NOTE: The Doppler effect is the phenomenon due to which there is a change in frequency or wavelength of a wave received by an observer who is moving relative to the wave source.

Let v_S and v_P be the velocities of S and P waves respectively. Let L be the distance between the epicentre and the seismograph such that

L = v_St_S ….....(i)

L = v_Pt_P …....(ii)

Where, t_S and t_P are the time taken by the S and P waves to reach the seismograph from the epicentre respectively.

It is given that v_P = 8 km/s

v_S = 4 km/s

From equations (i) and (ii),

v_S t_S = v_P t_P

⇒ 4t_S = 8 t_P

⇒ t_S = 2 t_P …..(iii)

It is also given that t_S – t_P = 4 min = 240 s

2t_P – t_P = 240

⇒ t_P = 240

And t_S = 2 × 240 = 480 s

From equation (ii),

L = 8 km/s × 240 s

⇒ L = 1920 km

Hence, the earthquake occurs at a distance of 1920 km from the seismograph.

Given,

Ultrasonic beep frequency emitted by the bat, f = 40kHz=40×10³ Hz

Velocity of the bat, v_b = 0.03v

where, v = velocity of sound in air

The apparent frequency of the sound striking the wall is given by

f’ = [v/(v-v_b)]f

⇒ f’ = [v/(v-0.03v)]f

⇒ f’ = f/0.97

⇒ f’ = (40×10³ Hz)/0.97

⇒ f’ = 41.237 × 10³ Hz

⇒ f’ = 41.237 kHz

This frequency (fs) is reflected back from the stationary wall. Let f’’ be the frequency received by the bat.

f’’ = [(v+v₀)/v]f’

⇒ f’’ = [(v+0.03v)/v]f’

⇒ f’’ = 1.03f’

⇒ f’’ = 1.03×41.237 kHz

⇒ f’’ = 42.474 kHz

NOTE: The Doppler effect is the phenomenon due to which there is a change in frequency or wavelength of a wave received by an observer who is moving relative to the wave source.