Simple Harmonic Motion

Question 1:

Question 2:

No. As motion is a change in position of an object with respect to time or a reference point, it is not an example of periodic motion.

Question 3:

No. The resultant force on the particle is maximum at the extreme positions.

Question 4:

Yes. Simple harmonic motion can take place in a non-inertial frame. However, the ratio of the force applied to the displacement cannot be constant because a non-inertial frame has some acceleration with respect to the inertial frame. Therefore, a fictitious force should be added to explain the motion.

Question 5:

No, we cannot say anything from the given information. To determine the displacement of the particle using its velocity at any instant, its mean position has to be known.

Question 6:

Yes, its shadow on a horizontal plane moves in simple harmonic motion. The projection of a uniform circular motion executes simple harmonic motion along its diameter (which is the shadow on the horizontal plane), with the mean position lying at the centre of the circle.

Question 7:

No. It does not make me unhappy because the number of times a particle crosses the mean and extreme positions does not depend on the speed of the particle.

Question 8:

The mean position of a particle executing simple harmonic motion is fixed, whereas its extreme position keeps on changing. Therefore, when we use stop watch to measure the time between consecutive passage, we are certain about the mean position.

Question 9:

Figure (a) shows the graph of the applied force against the position of the particle.

(a)


(b)


(c)

Question 10:

No. It cannot be negative because the minimum potential energy of a particle executing simple harmonic motion at mean position is zero. The potential energy increases in positive direction at the extreme position.

However, if we choose zero potential energy at some other point, say extreme position, the potential energy can be negative at the mean position.

Question 11:

Statement A is more appropriate because the energy of a system in simple harmonic motion is given by $E=\frac{1}{2}m {\omega }^2{A}^2.$

If the mass (m) and angular frequency (ω) are made constant, Energy (E) becomes proportional to the square of amplitude (A²).
i.e. E ∝ A²

Therefore, according to the relation, energy increases as the amplitude increases.

Question 12:

According to the relation:
$T=2\pi \sqrt{\frac{l}{g}}$

The time period (T) of the pendulum becomes proportional to the square root of inverse of g if the length of the pendulum is kept constant.
i.e. $T\propto \sqrt{\frac{1}{g}}$

Also, acceleration due to gravity (g) at the poles is more than that at equator. Therefore, the time period decreases and the clock gains time.

Question 13:

No. According to the relation:
$T=2\pi \sqrt{\frac{l}{g}}$
The time period of the pendulum clock depends upon the acceleration due to gravity. As the earth-satellite is a free falling body and its g_effective (effective acceleration due to gravity ) is zero at the satellite, the time period of the clock is infinite.

Question 14:

The time period of a pendulum depends on the length and is given by the relation, $T=2\pi \sqrt{\frac{l}{g}}$.
As the effective length of the pendulum first increases and then decreases, the time period first increases and then decreases.

Question 15:

Yes.

Time period of a spring mass system is given by,
$T=2\pi \sqrt{\frac{m}{k}}$                        ...(1)              .
where m is mass of the block, and
           k  is the spring constant

Time period is also given by the relation,
 $T=2\pi \sqrt{\frac{x_0}{g}}$                       ...(2)
where, x₀ is extension of the spring, and
            g is acceleration due to gravity

From the equations (1) and (2), we have:
 $mg=k{x}_0$

$\Rightarrow k=\frac{mg}{x_0}$

Substituting the value of k in the above equation, we get:
$T=2\pi \sqrt{\frac{m}{{\displaystyle \frac{mg}{x_0}}}}=2\mathrm{\pi }\sqrt{\frac{{\mathrm{x}}_0}{\mathrm{g}}}$
Thus, we can find the time period if the value of extension x₀ is known.

Question 16:

When the frequency of soldiers' feet movement becomes equal to the natural frequency of the bridge, and resonance occurs between soldiers' feet movement and movement of the bridge, maximum transfer of energy occurs from soldiers' feet to the bridge, which increases the amplitude of vibration. A continued increase in the amplitude of vibration, however, may lead to collapsing of the bridge.

Question 1:

As the observer moves with a constant velocity along the same axis, he sees the same force on the particle and finds the motion of the particle is not simple harmonic motion.

Question 2:

(a) As x increases k increases.

A body is said to be in simple harmonic motion only when,
   F = $-$kx                          ...(1)
where F is force,
           k is force constant, and
           x is displacement of the body from the mean position.

Given:
  F = -k$\sqrt{x}$                         ...(2)

On comparing the equations (1) and (2), it can be said that in order to execute simple harmonic motion, k should be proportional to $\sqrt{x}$ . Thus, as x increases k increases.

Question 3:

(b) an extreme position

One oscillation is said to be completed when the particle returns to the extreme position i.e. from where it started.

Question 4:

(a) v_max

Because the time period of a simple harmonic motion is defined as the time taken to complete one oscillation_.

Question 5:

(d) zero

Displacement is defined as the distance between the starting and the end point through a straight line. In one complete oscillation, the net displacement is zero as the particle returns to its initial position.

Question 6:

(c) 4A

In an oscillation, the particle goes from one extreme position to other extreme position that lies on the other side of mean position and then returns back to the initial extreme position. Thus, total distance moved by particle is,
2A + 2A = 4A.

Question 7:

(d) zero

The acceleration changes its direction (to opposite direction) after every half oscillation. Thus, net acceleration is given as,
Aω²+ ( -Aω²) = 0

Question 8:

(d) simple harmonic with amplitude  $\sqrt{A^2+B^2}$

x = A sin ωt + B cos ωt                                           ...(1)

 $$\begin{gathered} \mathrm{Acceleration}, \\ a=\frac{d^{2}x}{d{\mathrm{t}}^2}=\frac{d^2}{d{\mathrm{t}}^2}(A \sin \omega t + B \cos \omega t) \\ =\frac{\mathrm{d}}{\mathrm{dt}}(A\omega \cos \omega t - B\omega \sin \omega t) \\ =-A{\omega }^2 \mathrm{sin\omega t} - B{\omega }^2 \cos \mathrm{\omega t} \\ =-{\omega }^2(A \sin \omega t\mathit{ }+ B \cos \omega t ) \\ = -{\omega }^{2}x \end{gathered}$$

For a body to undergo simple harmonic motion,
acceleration, a = $-$kx.                                        ...(2)
Therefore, from the equations (1) and (2), it can be seen that the given body undergoes simple harmonic motion with amplitude,  A $=\sqrt{A^2+B^2}$.2

Question 9:

(c) on a circle

We know, $\frac{d^2}{d{t}^2}\overrightarrow{r}=-{\omega }^2 \overrightarrow{r}$
But there is a phase difference of 90^o between the x and y components because of which the particle executes a circular motion and hence, the projection of the particle on the diameter executes a simple harmonic motion.

Question 10:

(b) B

 At t = 0,
 Displacement $\left(x_0\right)$ is given by,   
 x₀ = A + sin ω(0) = A

 Displacement x will be maximum when sinωt is 1
 or,
 x_m = A + B

Amplitude will be:
x_m$-$ x_o = A + B $-$ A = B

Question 11:

(c) phase

Because the direction of motion of particles A and B is just opposite to each other.

Question 12:

 (d) remain E
Mechanical energy (E) of a spring-mass system in simple harmonic motion is given by,
$E_{}=\frac{1}{2}m{\omega }^2{A}^2$
where m is mass of body, and
           $\omega$ is angular frequency.

Let m₁ be the mass of the other particle and ω₁ be its angular frequency.
New angular frequency ω₁ is given by,
${\omega }_1=\sqrt{\frac{k}{m_1}}=\sqrt{\frac{k}{2m}} (m_1=2m)$

New energy E₁ is given as,
$$\begin{gathered} E_1 = \frac{1}{2}{m}_1{\omega }_1^2{A}^2 \\ =\frac{1}{2}\left(2m\right)(\sqrt{\frac{k}{2m}}{)}^2{A}^2 \\ =\frac{1}{2}m{\omega }^2{A}^2 =E \end{gathered}$$

Question 13:

(a)$\frac{1}{2}m{\omega }^2{A}^2$

It is the total energy in simple harmonic motion in one time period.

Question 14:

(c) 2v

Because in one complete oscillation, the kinetic energy changes its value from zero to maximum, twice.

Question 15:

 (d) become T/$\sqrt{2}$T/2√

Time period $\left(T\right)$ is given by,
$T = 2\mathrm{\pi }\sqrt{\frac{m}{k}}$
where m is the mass, and
          k is spring constant.

When the spring is divided into two parts, the new spring constant k₁ is given as,
k₁ = $2k$

New time period T₁:
T₁  = $2\mathrm{\pi }\sqrt{\frac{m}{2k}}=\frac{1}{\sqrt{2}}2\mathrm{\pi }\sqrt{\frac{m}{k}}=\frac{1}{\sqrt{2}}\mathrm{T}$

Question 16:

(d)$\sqrt{\frac{k_2}{k_1}}$

Maximum velocity, v = Aω
where A is amplitude and ω is the angular frequency.

Further, ω = $\sqrt{\frac{k}{m}}$
Let A and B be the amplitudes of particles A and B respectively. As the maximum velocity of particles are equal,

$$\begin{gathered} i.e. v_A=v_B \\ \mathrm{or}, \\ A{\omega }_A=B{\omega }_B \\ \Rightarrow A\sqrt{\frac{k_1}{m_A}}= B\sqrt{\frac{k_2}{m_B}} \\ \Rightarrow A\sqrt{\frac{k_1}{m}}= B\sqrt{\frac{k_2}{m}} (m_A=m_B=m) \\ \Rightarrow \frac{A}{B}= \sqrt{\frac{k_2}{k_1}} \end{gathered}$$

Question 17:

(c) remain same

Because the frequency ($\nu =\frac{1}{2\pi }\sqrt{\frac{k}{m}}$) of the system is independent of the acceleration of the system.

Question 18:

(c) remain same

As the frequency of the system is independent of the acceleration of the system.

Question 19:

(d) $\sqrt{6}$ times slower

The acceleration due to gravity at moon is g/6.

Time period of pendulum is given by, $T = 2\pi \sqrt{\frac{l}{g}}$

Therefore, on moon, time period will be:

T_moon = $2\pi \sqrt{\frac{l}{g_{moon}}}=2\pi \sqrt{\frac{l}{\left({\displaystyle \raisebox{1ex}{$g$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}\right)}}=\sqrt{6}\left(2\pi \sqrt{\frac{l}{g}}\right)=\sqrt{6}T$

Question 20:

(d) give correct time

Because the time period of a spring-mass system does not depend on the acceleration due to gravity.

Question 21:

(c) its length should be decreased to keep correct time

Time period of pendulum,
T = $2\pi \sqrt{\frac{l}{g}}$
At higher altitudes, the value of acceleration due to gravity decreases. Therefore, the length of the pendulum should be decreased to compensate for the decrease in the value of acceleration due to gravity.

Question 1:

(c) will go in a circular path

As the acceleration due to gravity acting on the bob of pendulum, due to free fall gives a torque to the pendulum, the bob goes in a circular path.

Question 2:

(a) A simple harmonic motion is necessarily periodic.
(b) A simple harmonic motion is necessarily oscillatory.

A periodic motion need not be necessarily oscillatory. For example, the moon revolving around the earth.
Also, an oscillatory motion need not be necessarily periodic. For example, damped harmonic motion.

Question 3:

(a) periodic

Because the particle covers one rotation after a fixed interval of time but does not oscillate around a mean position.

Question 4:

(a) periodic

Because the particle completes one rotation in a fixed interval of time but does not oscillate around a mean position.

Question 5:

(d) none of them

As the particle does not complete one rotation in a fixed interval of time, neither does it oscillate around a mean position.

Question 6:

(a) periodic
(b) oscillatory
(d) angular simple harmonic

Because it completes one oscillation in a fixed interval of time and the oscillations are in terms of rotation of the body through some angle.

Question 7:

$$\begin{gathered} \left(c\right) \stackrel{\rightharpoonup }{a}.\stackrel{\rightharpoonup }{r} \\ \left(d\right) \stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{r} \end{gathered}$$

In S.H.M.,
F = -kx
Therefore, $\stackrel{\rightharpoonup }{F.}\stackrel{\rightharpoonup }{r}$ will always be negative. As acceleration has the same direction as the force, $\stackrel{\rightharpoonup }{a}.\stackrel{\rightharpoonup }{r}$ will also be negative, always.

Question 8:

$\left(a\right) \stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{a}$

As the direction of force and acceleration are always same, $\stackrel{\rightharpoonup }{F}.\stackrel{\rightharpoonup }{a}$is always positive.

Question 9:

$$\begin{gathered} \left(a\right)\stackrel{\rightharpoonup }{F} x \stackrel{\rightharpoonup }{a} \\ \left(b\right)\stackrel{\rightharpoonup }{v} x \stackrel{\rightharpoonup }{\stackrel{\rightharpoonup }{r}} \\ \left(c\right)\stackrel{\rightharpoonup }{a}x \stackrel{\rightharpoonup }{r} \\ \left(d\right)\stackrel{\rightharpoonup }{F} x \stackrel{\rightharpoonup }{r} \end{gathered}$$

As $\stackrel{\rightharpoonup }{F}, \stackrel{\rightharpoonup }{a}, \stackrel{\rightharpoonup }{r} and \stackrel{\rightharpoonup }{v}$ are either parallel or anti-parallel to each other, their cross products will always be zero.

Question 1:

(c) on a straight line
(d) periodic

If the particle were dropped from the surface of the earth, the motion of the particle would be SHM. But when it is dropped from a height h, the motion of the particle is not SHM because there is no horizontal velocity imparted. In that case, the motion of the particle would be periodic and in a straight line.

Question 2:

It is given,
Amplitude of the simple harmonic motion, A =10 cm

At t = 0 and  x = 5 cm,
Time period of the simple harmonic motion, T  = 6 s

Angular frequency (ω) is given by,
    $\omega =\frac{2\mathrm{\pi }}{\mathrm{T}}=\frac{2\mathrm{\pi }}{6}=\frac{\mathrm{\pi }}{3} {\sec }^{-1}$

Consider the equation of motion of S.H.M,
     Y = Asin $\left(\omega t+\varphi \right)$                                 ...(1)
where Y is displacement of the particle, and
           $\varphi$ is phase of the particle.

On substituting the values of A, t and ω in equation (1), we get:
     5 = 10sin(ω × 0 + Ï•)   
$\Rightarrow$5 = 10sin Ï•

     $$\begin{gathered} \sin \mathrm{\varphi }=\frac{1}{2} \\ \Rightarrow \mathrm{\varphi }=\frac{\mathrm{\pi }}{6} \end{gathered}$$

∴ Equation of displacement can be written as,
     $x=\left(10 \mathrm{cm}\right) \sin \left(\frac{\mathrm{\pi }}{3}t+\frac{\mathrm{\pi }}{6}\right)$

(ii) At t = 4 s,
     $$\begin{gathered} x=10\sin \left[\frac{\mathrm{\pi }}{3}4+\frac{\mathrm{\pi }}{6}\right] \\ =10\sin \left[\frac{8\mathrm{\pi }+\mathrm{\pi }}{6}\right] \\ =10\sin \left(\frac{9\mathrm{\pi }}{6}\right) \\ =10\sin \left(\frac{3\mathrm{\pi }}{2}\right) \\ = 10\sin \left(\pi +\frac{\mathrm{\pi }}{2}\right) \\ =-10\sin \frac{\mathrm{\pi }}{2}=-10 \end{gathered}$$

Acceleration is given by,
    a = −ω²x
      $$\begin{gathered} =\left(\frac{-{\mathrm{\pi }}^2}{9}\right)\times \left(-10\right) \\ =10.9\approx 11 \mathrm{cm}/{\sec }^{-2} \end{gathered}$$

Question 3:

It is given that:
Position of the particle, x = 2 cm = 0.02 m
Velocity of the particle, v = 1 ms^−1.
Acceleration of the particle, a = 10 ms^−2.

Let $\omega$ be the angular frequency of the particle.
The acceleration of the particle is given by,
     a = ω²x
$$\begin{gathered} \Rightarrow \omega =\sqrt{\frac{a}{x}}=\sqrt{\frac{10}{0.02}} \\ =\sqrt{500}=10\sqrt{5} \mathrm{Hz} \\ \mathrm{Time} \mathrm{period} \mathrm{of} \mathrm{the} \mathrm{motion} \mathrm{is} \mathrm{given} \mathrm{as}, \\ T=\frac{2\mathrm{\pi }}{\mathrm{\omega }}=\frac{2\mathrm{\pi }}{10\sqrt{5}} \\ =\frac{2\times 3.14}{10\times 2.236} \\ =0.28 \mathrm{s} \end{gathered}$$

Now, the amplitude A is calculated as,
     $$\begin{gathered} v=\omega \sqrt{A^2-x^2} \\ \Rightarrow v^2={\omega }^2 \left(A^2-x^2\right) \\ 1=500\left(A^2-0.0004\right) \\ \Rightarrow A=0.0489=0.049 \mathrm{m} \\ \Rightarrow A= 4.9 \mathrm{cm} \end{gathered}$$

Question 4:

It is given that:
Amplitude of the particle executing simple harmonic motion, A = 10 cm

To determine the distance from the mean position, where the kinetic energy of the particle is equal to its potential energy:
Let y be displacement of the particle,
      $\omega$ be the angular speed of the particle, and
      A be the amplitude of the simple harmonic motion.

Equating the mathematical expressions for K.E. and P.E. of the particle, we get:
$\left(\frac{1}{2}\right)m{\mathrm{\omega }}^2 \left(A^2-y^2\right)=\left(\frac{1}{2}\right)m{\mathrm{\omega }}^2{y}^2$
                A² − y² = y²
                      2y² = A²

$\Rightarrow y=\frac{A}{\sqrt{2}}=\frac{10}{\sqrt{2}}=5\sqrt{2}$
The kinetic energy and potential energy of the particle are equal at a distance of $5\sqrt{2}$ cm from the mean position.

Question 5:

It is given that:
Maximum speed of the particle, $v_{Max}$ = 10 cms^$-1$
Maximum acceleration of the particle, $a_{Max}$ = 50 cms⁻²â€‹

The maximum velocity of a particle executing simple harmonic motion is given by,
    $v_{Max}=A\omega$
where $\omega \mathrm{is} \mathrm{angular} \mathrm{frequency}, \mathrm{and}$
           A is amplitude of the particle.

Substituting the value of $v_{Max}$ in the above expression, we get:
      Aω = 10                           $...\left(1\right)$
   $\Rightarrow {\mathrm{\omega }}^2=\frac{100}{A^2}$

   a_Max = ω²A = 50 cms⁻¹
  $$\begin{gathered} \Rightarrow {\omega }^2=\frac{50}{A} ...\left(2\right) \\ \mathrm{From} \mathrm{the} \mathrm{equations} \left(1\right) \mathrm{and} \left(2\right), \mathrm{we} \mathrm{get}: \\ \frac{100}{A^2}=\frac{50}{A} \\ \Rightarrow A=2 \mathrm{cm} \\ \mathbf{\therefore } \omega =\sqrt{\frac{100}{A^2}}=5 {\sec }^{-1} \end{gathered}$$

To determine the positions where the speed of the particle is 8 ms^-1, we may use the following formula:
      v² = ω² (A² − y²)
      where y is distance of particle from the mean position, and
                 v is velocity of the particle.

On substituting the given values in the above equation, we get:
      64 = 25 (4 − y²)
$\Rightarrow \frac{64}{25}=4-y^2$

⇒ 4 − y² = 2.56
⇒       y² = 1.44
⇒​       y  = $\sqrt{1.44}$
⇒        y = ± 1.2 cm   (from the mean position)

Question 6:

Given:
Equation of motion of the particle executing S.H.M.,
$$\begin{gathered} x=\left(2.0 \mathrm{cm}\right) \sin \left[\left(100 s^{-1}\right)t+\frac{\mathrm{\pi }}{6}\right] \\ \mathrm{Mass} \mathrm{of} \mathrm{the} \mathrm{particle}, m=10 \mathrm{g} \end{gathered}$$                 ...(1)
General equation of the particle is given by,
$x=A\sin (\omega t+\varphi )$                                        ...(2)

On comparing the equations (1) and (2) we get:

(a) Amplitude, A is 2 cm.
     Angular frequency, ω is 100 s⁻¹â€‹.

     $$\begin{gathered} \mathrm{Time} \mathrm{period} \mathrm{is} \mathrm{calculated} \mathrm{as}, \\ T=\frac{2\mathrm{\pi }}{\omega }=\frac{2\mathrm{\pi }}{100}=\frac{\mathrm{\pi }}{50}\mathrm{s} \\ =0.063 \mathrm{s} \end{gathered}$$

Also, we know -
          $$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{m}{k}} \\ \mathrm{where} \mathrm{k} \mathrm{is} \mathrm{the} \mathrm{sprin}g \mathrm{constant}. \\ \Rightarrow T^2=4{\mathrm{\pi }}^2\frac{m}{k} \\ \Rightarrow k=\frac{4{\mathrm{\pi }}^{2}m}{T^2}={10}^5 \mathrm{dyne}/\mathrm{cm} \\ =100 \mathrm{N}/\mathrm{m} \end{gathered}$$


(b) At t = 0 and x = 2 cm $\sin \frac{\mathrm{\pi }}{6}$
                           $=2\times \frac{1}{2}=1 \mathrm{cm} \mathrm{from} \mathrm{the} \mathrm{mean} \mathrm{position},$

    We know:     
     x = A sin (ωt + Ï•)

   Using $v=\frac{dx}{dt},$ we get:
            v = Aω cos (ωt + Ï•)
              $$\begin{gathered} =2\times 100 \cos \left(0+\frac{\mathrm{\pi }}{6}\right) \\ =200\times \frac{\sqrt{3}}{2} \\ =100\sqrt{3} {\mathrm{cms}}^{-1} \\ =1.73 {\mathrm{ms}}^{-1} \end{gathered}$$


(c) Acceleration of the particle is given by,
       a = $-{\omega }^2$x
          = 100²×1 = 10000 cm/s²

Question 7:

Given:
The equation of motion of a particle executing S.H.M. is,
$x=5 \sin \left(20t+\frac{\mathrm{\pi }}{3}\right)$ 
 
The general equation of S..H.M. is give by,
 $x=A \sin (\omega t+\varphi )$  

(a) Maximum displacement from the mean position is equal to the amplitude of the particle. 
     As the velocity of the particle is zero at extreme position, it is at rest.
    $\therefore \mathrm{Displacement}$ x = 5, which is also the amplitude of the particle.
    $$\begin{gathered} \Rightarrow 5=5 \sin \left(20t+\frac{\mathrm{\pi }}{3}\right) \\ \mathrm{Now}, \\ \sin \left(20t+\frac{\mathrm{\pi }}{3}\right)=1=\sin \frac{\mathrm{\pi }}{2} \\ \Rightarrow 20t+\frac{\mathrm{\pi }}{3}=\frac{\mathrm{\pi }}{2} \\ \Rightarrow t=\frac{\mathrm{\pi }}{120} \mathrm{s} \end{gathered}$$

The particle will come to rest at $\frac{\mathrm{\pi }}{120} \mathrm{s}$

(b)  Acceleration is given as,
     a = ω²x
        $={\mathrm{\omega }}^2\left[5 \sin \left(20t+\frac{\mathrm{\pi }}{3}\right)\right]$
 
    For a = 0,
    $$\begin{gathered} 5 \sin \left(20t+\frac{\mathrm{\pi }}{3}\right)=0 \\ \Rightarrow \sin \left(20t+\frac{\mathrm{\pi }}{3}\right)=\sin \pi \\ \Rightarrow 20t=\mathrm{\pi }-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow t=\frac{\mathrm{\pi }}{30} \mathrm{s} \end{gathered}$$

(c) The maximum speed $\left(v\right)$ is given by,
     $v=\mathrm{A}\omega \cos \left(\omega t+\frac{\mathrm{\pi }}{3}\right)$                          (using $v=\frac{dx}{dt}$)
      $=20\times 5 \cos \left(20t+\frac{\mathrm{\pi }}{3}\right)$
     For maximum velocity:
     $$\begin{gathered} \cos \left(20t+\frac{\mathrm{\pi }}{3}\right)=-1= \cos \mathrm{\pi } \\ \Rightarrow 20t=\mathrm{\pi }-\frac{\mathrm{\pi }}{3}=\frac{2\mathrm{\pi }}{3} \\ \Rightarrow t=\frac{\mathrm{\pi }}{30} \mathrm{s} \end{gathered}$$

Question 8:

It is given that a particle executes S.H.M.
Equation of S.H.M. of the particle:
x = 2.0 cos (50$\mathrm{\pi }$t + tan⁻¹0.75)
   = 2.0 cos (50$\mathrm{\pi }$t + 0.643)

(a) Velocity of the particle is given by,
     $v=\frac{dx}{dt}$
     v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643)

   As the particle comes to rest, its velocity becomes be zero.
   ⇒​ v = −100$\mathrm{\pi }$ sin (50$\mathrm{\pi }$t + 0.643) = 0
   ⇒                   sin (50$\mathrm{\pi }$t + 0.643) = 0 = sin $\mathrm{\pi }$

   When the particle initially comes to rest,
    50$\mathrm{\pi }$t + 0.643 = $\pi$
   ⇒                 t = 1.6 × 10⁻²s

(b) Acceleration is given by,
     $$\begin{gathered} a=\frac{dv}{dt} \\ =-100\mathrm{\pi }\times 50\pi \cos \left(50\mathrm{\pi }t+0.643\right) \end{gathered}$$

     For maximum acceleration:
     cos (50$\mathrm{\pi }$t + 0.643) = −1 = cos $\mathrm{\pi }$ (max)             (so that a is max)
⇒                             t = 1.6 ×  10⁻² s

(c) When the particle comes to rest for the second time, the time is given as,
     50$\mathrm{\pi }$t + 0.643 = 2$\mathrm{\pi }$
⇒ â€‹                    t = 3.6 × 10⁻² s

Question 9:

As per the conditions given in the question,
$$\begin{gathered} y_1=\frac{A}{2}; \\ {y}_2=A \end{gathered}$$    (for the given two positions)

Let y₁and y₂ be the displacements at the two positions and A be the amplitude.
Equation of motion for the displacement at the first position is given by,
 y₁ = Asinωt₁
As displacement is equal to the half of the amplitude,
        $\frac{A}{2}=A \sin \omega t_1$
$$\begin{gathered} \Rightarrow \sin \omega t_1=\frac{1}{2} \\ \Rightarrow \frac{2\mathrm{\pi }\times t_1}{T}=\frac{\mathrm{\pi }}{6} \\ \Rightarrow t_1=\frac{T}{12} \end{gathered}$$


The displacement at second position is given by,
y₂ = A sin ωt₂
As displacement is equal to the amplitude at this position,
⇒        A = A sin ωt₂
⇒ sinωt₂ = 1
$$\begin{gathered} \Rightarrow \omega t_2=\frac{\mathrm{\pi }}{2} \\ \Rightarrow \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\right)t_2=\frac{\mathrm{\pi }}{2} \left(\because \sin \frac{\mathrm{\pi }}{2}=1\right) \\ \Rightarrow t_2=\frac{T}{4} \\ \therefore t_2-t_1=\frac{T}{4}-\frac{T}{12}=\frac{T}{6} \end{gathered}$$

Question 10:

Given:
Spring constant, k =0.1 N/m
Time period of the pendulum of clock, T = 2 s
Mass attached to the string, m, is to be found.

The relation between time period and spring constant is given as,
$\mathrm{T}=2\mathrm{\pi } \sqrt{\left(\frac{m}{k}\right)}$
On substituting the respective values, we get:
$$\begin{gathered} 2=2\mathrm{\pi }\sqrt{\frac{m}{k}} \\ \Rightarrow {\mathrm{\pi }}^2\left(\frac{m}{0.1}\right)=1 \\ \therefore m=\frac{0.1}{{\mathrm{\pi }}^2}=\frac{0.1}{10} \\ =0.01 \mathrm{kg}\approx 10 \mathrm{g} \end{gathered}$$

Question 11:

An equivalent simple pendulum has same time period as that of the spring mass system.
The time period of a simple pendulum is given by,
$T_{\mathrm{p}}=2\mathrm{\pi }\sqrt{\left(\frac{l}{g}\right)}$
where l is the length of the pendulum, and
           g is acceleration due to gravity.

Time period of the spring is given by,
$T_{\mathrm{s}}=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$
where m is the mass, and 
           k is the spring constant.

Let x be the extension of the spring.
For small frequency, T_P â€‹can be taken as equal to T_S.    



$$\begin{gathered} \Rightarrow \sqrt{\left(\frac{l}{g}\right)}=\sqrt{\left(\frac{m}{k}\right)} \\ \Rightarrow \left(\frac{l}{g}\right)=\left(\frac{m}{k}\right) \\ \Rightarrow l=\frac{mg}{k}=\frac{F}{k}=x \end{gathered}$$
($\because$ restoring force = weight = mg) 

$\therefore$l = x (proved)

Question 12:

It is given that:
Amplitude of simple harmonic motion, x = 0.1 m
Time period of simple harmonic motion, T  = 0.314 s
Mass of the block, m = 0.5 kg
Weight of the block, W = mg = 0.5$\times$10 = 5 kg          $\left(\because g=10 {\mathrm{ms}}^{-2}\right)$

Total force exerted on the block = Weight of the block + spring force



Periodic time of spring is given by,
$$\begin{gathered} \mathrm{T}=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)} \\ \Rightarrow 0.314=2\mathrm{\pi }\sqrt{\left(\frac{0.5}{k}\right)} \\ \Rightarrow k=200 \mathrm{N}/\mathrm{m} \end{gathered}$$

∴ The force exerted by the spring on the block $\left(F\right)$ is,
    F = kx = 200.0 × 0.1 = 20 N

    Maximum force = F + weight of the block
                             = 20 + 5 = 25 N

Question 10:

It is given that:
Mass of the body, m = 2 kg
Time period of the spring mass system, T = 4 s
The time period for spring-mass system is given as,
$T=2\mathrm{\pi }\sqrt{\left(\frac{m}{k}\right)}$
where k is the spring constant.

On substituting the respective values, we get:
$$\begin{gathered} \Rightarrow 4=2\mathrm{\pi }\sqrt{\frac{2}{k}} \\ \Rightarrow 2=\mathrm{\pi }\sqrt{\frac{2}{k}} \\ \Rightarrow 4={\mathrm{\pi }}^2\left(\frac{2}{k}\right) \\ \Rightarrow k=\frac{2{\mathrm{\pi }}^2}{4} \\ =\frac{{\mathrm{\pi }}^2}{2}= 5 \mathrm{N}/\mathrm{m} \end{gathered}$$

As the restoring force is balanced by the weight, we can write:
        F = mg = kx
    $\Rightarrow x=\frac{mg}{k}=\frac{2\times 10}{5}=4$

∴ Potential Energy $\left(U\right)$ of the spring is,
      $$\begin{gathered} U=\left(\frac{1}{2}\right)k{x}^2=\frac{1}{2}\times 5\times 16 \\ =5\times 8=40 \mathrm{J} \end{gathered}$$                 

Question 11:

(a) displacement from the mean position

For S.H.M.,
F = -kx
ma = - kx                  (F = ma)
or,
a = $-\frac{k}{m}x$
Thus, acceleration is proportional to the displacement from the mean position but in opposite direction.

Question 12:

(a) on a straight line
(c) periodic
(d) simple harmonic

The given equation is a solution to the equation of simple harmonic motion. The amplitude is $(\stackrel{\rightharpoonup }{i}+2\stackrel{\rightharpoonup }{j})A$, following equation of straight line y = mx + c. Also, a simple harmonic motion is periodic.

Question 13:

(d) with time period $\frac{\pi }{\omega }$

Given equation:
x = x_osin²ωt

⇒​$x=\frac{x_0}{2}(\cos 2\omega t- 1)$

Now, the amplitude of the particle is x_o/2 and the angular frequency of the SHM is 2ω.

Thus, time period of the SHM = $\frac{2\pi }{\text{angular frequency}}=\frac{2\pi }{2\omega }=\frac{\pi }{\omega }$

Question 14:

(d) the average potential energy in one time period is equal to the average kinetic energy in this period.

The kinetic energy of the motion is given as,
$\frac{1}{2}k{A}^2 {\cos }^2 \omega t$ 

The potential energy is calculated as,
$\frac{1}{2}k{A}^2 {\sin }^2 \omega t$

As the average of the cosine and the sine function is equal to each other over the total time period of the functions, the average potential energy in one time period is equal to the average kinetic energy in this period.

Question 15:

(a) the maximum potential energy equals the maximum kinetic energy
(b) the minimum potential energy equals the minimum kinetic energy

In SHM,
maximum kinetic energy    = $\frac{1}{2}k{A}^2$ 
maximum potential energy = $\frac{1}{2}k{A}^2$

The minimum value of both kinetic and potential energy is zero. 
Therefore, in a simple harmonic motion the maximum kinetic energy and maximum potential energy are equal. Also, the minimum kinetic energy and the minimum potential energy are equal.

Question 16:

(a) the measured times are same
(b) the measured speeds are same

The effect of gravity on the object as well as on the pendulum clock is same in both cases; the time measured is also same. As the time measured is same, the speeds are same.

Question 16:

(a) A simple pendulum
(b) A physical pendulum

As the time period of a simple pendulum and a physical pendulum depends on the acceleration due the gravity, the time period of these pendulums changes when they are taken to the moon.