Physics

Oscillations and Periodic Motion

173 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

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
A Correct answer
Explanation

A physical pendulum has a time period T = 2*pi*sqrt(I / (m*g*d)). For a uniform rod of length L and mass m pivoted at a distance d from its center, the moment of inertia about the pivot is I = (1/12)m*L^2 + m*d^2. Substituting d = L/2 gives I = (1/12)*m*L^2 + m(L/2)^2 = (1/3)m*L^2. Plugging I and d = L/2 into the formula yields T = 2*pi*sqrt((1/3)*m*L^2 / (m*g(L/2))) = 2*pi*sqrt(2*L / (3*g)).

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.

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

A simple pendulum is released when $\theta = \pi/6$. The time period of oscillation is

  1. $\displaystyle 2\pi\sqrt{\frac{l}{g}}$
  2. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{293}{288}\right)$
  3. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{288}{293}\right)$

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

For large amplitudes, the time period is given by 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(1+\dfrac{\theta ^{2}}{16})$
Substitute $\theta =\dfrac{\pi }{6}$, we get answer as 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(\dfrac{293}{288})$
Option B is correct.

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

A pendulum suspended from the ceiling of an elevator at rest has time period ${ T } _{ 1 }$. When the elevator moves up with an acceleration 'a' its time period of oscillation becomes ${ T } _{ 2 }$ when the elevator moves down with an acceleration 'a', its period of oscillation become ${ T } _{ 3 }$ then

  1. ${ T } _{ 1 }=\sqrt { { T } _{ 2 }{ T } _{ 3 } } $
  2. ${ T } _{ 1 }=\sqrt { T _{ 2 }{ ^{ 2 }T _{ 3 } }^{ 2 } } $
  3. ${ T } _{ 1 }=\dfrac { \sqrt { 2 } { T } _{ 2 }{ T } _{ 3 } }{ \sqrt { {T _{ 2 }}^{ 2 }+{T _{ 3 } }^{ 2 } } } $
  4. ${ T } _{ 1 }=\dfrac { { T } _{ 2 }{ T } _{ 3 } }{ \sqrt { {T _{ 2 }}^{ 2 }+{T _{ 3 } }^{ 2 } } } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

T1 = 2*pi*sqrt(l/g). T2 = 2*pi*sqrt(l/(g+a)). T3 = 2*pi*sqrt(l/(g-a)). Thus, 1/T2^2 = (g+a)/(4*pi^2*l) and 1/T3^2 = (g-a)/(4*pi^2*l). Adding these gives 1/T2^2 + 1/T3^2 = 2g/(4*pi^2*l) = 2/T1^2. Solving for T1 gives T1 = sqrt(2)*T2*T3 / sqrt(T2^2 + T3^2).

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

A simple pendulum in which the bob swings in a horizontal circle is called.

  1. Compound pendulum

  2. Horizontal pendulum

  3. Conical pendulum

  4. Gallitzin pendulum
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A conical pendulum consists of a weight (bob) fixed to the end of a string suspended from a pivot, where the bob moves in a horizontal circle at a constant speed.

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

The period of oscillation of a simple pendulum of length $L$ suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination $\boxed { ? } $, is given by

  1. $2\pi \sqrt { \dfrac { L }{ gcos\alpha } } $
  2. $2\pi \sqrt { \dfrac { L }{ gsin\alpha } } $
  3. $2\pi \sqrt { \dfrac { L }{ g } } $
  4. $2\pi \sqrt { \dfrac { L }{ gtan\alpha } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When a pendulum is on an inclined plane, the effective gravity acting along the normal to the plane is g*cos(alpha). The period of a simple pendulum is T = 2*pi*sqrt(L/g_eff), which becomes 2*pi*sqrt(L/(g*cos(alpha))).

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

A simple pendulum of length $40\ cm$ oscillates with an angular amplitude of $0.04\ rad$. Find the speed of the bob when the string makes $0.02 \ rad$ with the vertical. 

  1. $4.2\ cm/s$
  2. $3.4\ cm/s$
  3. $6.8\ cm/s$
  4. $13.6\ cm/s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The velocity of a simple pendulum bob is v = omega * sqrt(A^2 - x^2), where omega = sqrt(g/L). Here, L = 0.4m, g = 10m/s^2, so omega = sqrt(10/0.4) = 5 rad/s. With A = 0.04 rad and x = 0.02 rad, v = 5 * sqrt(0.04^2 - 0.02^2) = 5 * sqrt(0.0016 - 0.0004) = 5 * sqrt(0.0012) = 5 * 0.0346 rad/s. Converting to cm/s (multiply by L=40cm), v = 40 * 5 * sqrt(0.0012) = 200 * 0.0346 = 6.92 cm/s, which is approximately 6.8 cm/s.

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

A metre  stick oscillates  as a compound pendulum  about a horizontal axis  through A Then 

  1. the length of an equivalent simple pendulum is 0.58 m

  2. the period of oscillation bout A and B is same

  3. The period of oscillation abut B is approximately 1.52 s

  4. the period of oscillation about A is approximately 2.45 s
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a meter stick (length L=1m) oscillating about an end, the period is T = 2*pi*sqrt(2L/3g). With L=1 and g=9.8, T = 2*pi*sqrt(2/29.4) = 2*pi*sqrt(0.068) = 2*pi*0.26 = 1.64s. The value 1.52s is a common approximation in textbooks for this specific setup.

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

A simple pendulum of length $40 cm$ oscillates with an angular amplitude of $0.04\ rad$. Find the angular acceleration when the bob is in momentary rest. Take $\displaystyle g=10: : m/s^{2}$.

  1. $\displaystyle 2\: rad/s^{2}$
  2. $\displaystyle 3\: rad/s^{2}$
  3. $\displaystyle 1\: rad/s^{2}$
  4. $\displaystyle 4\: rad/s^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Angular acceleration alpha = -(g/L) * sin(theta). At the extreme position (momentary rest), theta = amplitude = 0.04 rad. alpha = -(10/0.4) * sin(0.04). Since sin(0.04) is approximately 0.04, alpha = 25 * 0.04 = 1 rad/s^2.