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

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.

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

A pendulum bob has a speed of $ 3 $ $ \mathrm{ms}^{-1} $ at  its lowest position. The pendulum is$ 0.5$ $ \mathrm{m}  $ long. The speed of the bob, when the length makes an angle of $ 60^{\circ}  $ to the vertical will be $ (g=10 $ $ \left(n s^{-1}\right) $

  1. $3$ $ m s^{-1} $
  2. $
    1 / 3 \mathrm{ms}^{-1}
    $
  3. $
    1 / 2 m s^{-1}
    $
  4. $
    2 m s^{-1}
    $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Apply energy conservation theorem$,$ 
energy at lowest position of Bob $=$ energy$ ,$ when Bob makes $60°$ to the vertical 
$1/2 mv^2 = 1/2 mv₁^2 + mgl(1 - cos60°)$
Here $v$ is speed at Lowest position $, v₁$ is speed $,$ when it makes $60°$ with vertical and $l$ is length of pendulum $.$
$[$Actually, height of Bob $,$ when it makes $60°$ with vertical $= l(1 - cos60°)] $
$∴ v^2 = v₁^2 + 2gl(1 - cos60°)$ 
$3^2 = v₁^2 + 2 × 10 × 0.5 (1 - 1/2)$ 
$9 = v₁^2 + 5$ 
$v₁^2 = 4 ⇒v₁ = 2m/s $
So$,$ speed of Bob $= 2m/s$
Hence,
option $(D)$ is correct answer.
Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

Which of the following will change the time period as they are taken to moon?

  1. A simple pendulum

  2. A physical pendulum

  3. A torsional pendulum

  4. A spring-mass system
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

$(i)$ For simple pendulum $T = 2\pi\sqrt{L/g}$
$(ii)$ For physical pendulum $T = 2\pi\sqrt{I/mgL}$
So in both above case, time period is changed if they are taken to the moon.
$(iii)$ For torsional pendulum $T = 2\pi\sqrt{I/C}$
$(iv)$ For spring-mass system $T = 2\pi\sqrt{m/k}$

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

A simple pendulum of length L and having a bob of mass m is suspended in a car. The car is moving on a circular track of radius R with a uniform speed v. If the pendulum makes small oscillations in a radial direction about its equilibrium positions, its time period of oscillation is:

  1. $ T = 2 \pi \sqrt{\dfrac{L}{g}}$
  2. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}+ \dfrac{v^4}{R^2}}}$
  3. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}+ \dfrac{v^2}{R}}}$
  4. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}- \dfrac{v^4}{R^2}}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The period of the pendulum, $T=2\pi \sqrt{\cfrac{L}{a}}$ where $a=$ $\text{resultant acceleration}\=\sqrt{g^2+{(\cfrac{V^2}{R})}^2}\quad\quad\quad\quad [\cfrac{V^2}{R}=\text{centripital acceleratiop }, g=\text{acceleration due to gravity}]$.

$\therefore T=2\pi\sqrt{\cfrac{L}{g^2+\cfrac{V^4}{R^2}}}$
Option B is the correct answer.


Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

A pendulum beats seconds on the earth. Its time period on a stationary satellite of the earth will be

  1. Zero
  2. $1\ s$
  3. $2\ s$

  4. Infinity
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Inside a satellite, every object experiences weightlessness

Therefore Time period of a pendulum inside a satellite is $T= 2\pi \sqrt{\cfrac{L}{g}}$
as $g=0$
$\therefore T=\infty$ (Infinity)
A pendulum beats seconds on the earth. Its time period on a stationary satellite of the earth will be Infinity.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

Time period of simple pendulum in a satellite is

  1. Infinite

  2. Zero

  3. 2 sec

  4. Cannot be calculated
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Time period of simple pendulum is given by:
$\displaystyle T = 2\pi \sqrt{\frac{l}{g}}$
where l is the length of the pendulum.
Inside a satellite, $g = 0$
Hence, period will be infinite which means there will be no oscillation.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

The time period of a second's pendulum inside a satellite will be

  1. zero
  2. $1$ sec
  3. $2$ sec

  4. infinite
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

time period $\propto \sqrt { \cfrac { l }{ g }  } $ , as in the satellite there will be no gravity so the time period will be infinite.(logical explanation. : without gravity there will be no force on pendulum so it will not move a bit even in infinite time.)