Physics

Oscillations and Periodic Motion

162 Questions

Oscillations and periodic motion describe the movement of objects repeating their paths in regular intervals. Key concepts include simple pendulums, kinetic energy variations, and mechanical resonance. This physics topic is vital for various competitive exams.

Simple pendulumTime period calculationsKinetic energy in SHMMechanical resonanceDamped oscillations

Oscillations and Periodic Motion Questions

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Write the torque equation for the bob of a pendulum if it makes an angle of $\theta$ with the vertical and I is the moment of inertia of the bob w.r.t the point of suspension

  1. $I \dfrac{d^2 \theta}{dt^2}=mgL \cos \theta$
  2. $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$
  3. $I \dfrac{d^2 \theta}{dt^2}=mgL \tan \theta$
  4. $I \dfrac{d^2 \theta}{dt^2}=mg \sin \theta$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Taking the torque about the point of suspension, we can write $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$

The correct option is (b)

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A pendulum clock keeping correct time is taken to high altitudes,

  1. it will keep correct time

  2. its length should be increased to keep correct time

  3. its length should be decreased to keep correct time

  4. it cannot keep correct time even if the length is
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At high altitudes, g decreases. Since T = 2*pi*sqrt(l/g), T increases, meaning the clock runs slow. To keep correct time, T must be decreased, which requires decreasing the length l.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Restoring force on the bob of a simple pendulum of mass $100\ gm$ when its amplitude is ${ 1 }^{ 0 } $ is 

  1. $0.017\ N$
  2. $1.7\ N$
  3. $0.17\ N$
  4. $0.034\ N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The restoring force for a simple pendulum is F = mg*sin(theta). For theta = 1 degree, F = 0.1 * 9.8 * sin(1 degree) approx 0.1 * 9.8 * 0.01745 = 0.0171 N.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Simple pendulum of large length is made equal to the radius of earth. Its period of oscillation will be then?

  1. 83.5 minutes

  2. 59.8 minutes

  3. 42.3 minutes
    1. 15 minutes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The time period of simple pendulum is given by:

$T=2\pi \sqrt{\dfrac{L}{g}}$

When length of pendulum is equal to the radius of earth. $R=L=6371\,\,Km=6371\times {{10}^{3}}\,Km$

So, time is

$ T=2\pi \sqrt{\dfrac{6371\times {{10}^{3}}}{10}} $

$ T=2\pi \times 798.18 $

$ T=5012.604\,seconds $

$\therefore$ $ T=83.54\,minutes $

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

If the length of a clock pendulum increase by $0.2\%$ due to atmospheric temperature rise, then the loss in time of clock per day is 

  1. $86.4$s
  2. $43.2$s
  3. $72.5$s
  4. $32.5$s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$T=2\pi \sqrt{\dfrac{l}{g}}$

$\dfrac{\Delta T}{T}\times 100 = \dfrac{1}{2}\dfrac{\Delta l}{l}\times 100 $

$\dfrac{\Delta T}{24 \times 3600}\times 100 = \dfrac{1}{2}\dfrac{0.2}{100}\times 100 $

$\Delta T=86.4s$

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Find the time period of oscillations of a torsional pendulum, if the torsional constant of the wire is K = 10$\pi^2$J/rad. The moment of inertia of rigid body is 10 Kg m$^2$ about the axis of rotation.

  1. 2 sec

  2. 4 sec

  3. 16 sec

  4. 8 sec
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Time period of a torsional pendulum is given by
$T = 2 \pi \sqrt{\dfrac{I}{k}}$
$\Rightarrow T=2 \pi \sqrt{\dfrac{10}{10\pi^2}}=2 sec$

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A clock which has a pendulum made of brass keep correct time at ${30^0}C$? How many seconds it will gain or lose in day if the temperature falls to ${0^0}C$.

  1. It will lose $23.32$ sec per day
  2. It will gain $23.32$ sec per day
  3. It will lose $50$ sec per day
  4. It will gain $50$ sec per day
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The change in time period is delta_T = (1/2)*alpha*T*delta_theta. For brass, alpha is approx 1.8e-5. delta_T/T = (1/2)*alpha*delta_theta. The time lost/gained per day is (delta_T/T) * 86400. Plugging in values gives approx 23.3 seconds.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A simple pendulum of length $1$m has a bob of mass $100$g. It is displaced through an angle of $60^o$ from the vertical and then released . Find out K.E. of bob when it passes through mean position.

  1. $0.12$J
  2. $0.24$J
  3. $0.36$J
  4. $0.55$J
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Length of simple pendulum $=1m$

Mass $=1w\;gm=0.1kg$
It is displaced through as angle of $60$ for vertical 
Height of pendulum at starting position $=$ length $(1-ws\;60)$
                                                                   $=1\left( 1-0.5\right)$
                                                                   $=0.5m$
Potential energy $=mgh=0.1\times 10\times 0.5 =0.55$
when it is released and it reaches mean position its potential energy at starting point is converted to kinetic energy.
so K.E. of bob at mean position $=0.55.$
Hence, the answer is $0.55.$

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

A pendulum is formed by pivoting a long thin rod of length L and mass m about a point P on the rod which is a distance d above the center of the rod as shown. 
Now answer the following questions. 
1. The time period of this pendulum when d = L/2 will be

  1. $2\pi\sqrt { \dfrac { 2\ell }{ 3g }}$
  2. $2\pi\sqrt { \dfrac { 3\ell }{ 2g }}$
  3. $4\pi\sqrt { \dfrac { \ell }{ 3g }}$
  4. $\dfrac {2\pi} {3} \sqrt { \dfrac { 2\ell }{ g }}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The bob cf a simple pendulum is a spherical hollowe bal filled with water A pluyged hole near the bouthmol th oscilloting bob gets suddenly unplugged. During observation, till water is coming out, the time period of would 

  1. First increase and then decrease to the original value

  2. first decrease and then increase to the original value

  3. remain unchanged

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

As water leaks out, the center of mass of the bob initially moves downwards, increasing the effective length of the pendulum, which increases the time period. Once the water is empty, the center of mass returns to the center of the sphere, decreasing the period back to the original value.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

The bob of a simple pendulum executes  $S H M$  in water with a period  $t,$  while the period of oscillation of the bob is  $t _{ 0 }$  in air. Neglecting the frictional force of water and given that the density of the bob is  $( 4 / 3 ) \times 1000 kg / { m } ^ { 3 }.$  What relationship between  $t$  and  $t _ { 0 }$  is true ?

  1. $t = t _ { 0 }$
  2. $t = 4 t _ { 0 }$
  3. $t = 2 t _ { 0 }$
  4. $t = t _ { 0 } / 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The effective gravity in water is g' = g(1 - rho_water/rho_bob). Given rho_bob = 4/3 * 1000 and rho_water = 1000, g' = g(1 - 3/4) = g/4. Since T is proportional to 1/sqrt(g), T_water = T_air / sqrt(1/4) = 2 * T_air.