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 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.)

Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The period of oscillation $T$ of a pendulum of length $l$ at a place of acceleration due to gravity $g$ is given by $T=2\pi \sqrt {\dfrac {l}{g}}$. If the calculated length is $0.992$ times the actual length and if the value assumed for $g$ is $1.002$ times its actual value, the relative error in the computed value of $T$ is

  1. $0.005$
  2. $-0.005$
  3. $0.003$
  4. $-0.003$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Relative error will calculate by
$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{\Delta L}{L}-\dfrac{\Delta g}{g}\right]$

$\dfrac{\Delta T}{T}=\dfrac{1}{2}\left[\dfrac{0.992L}{L}-\dfrac{1.002 g}{g}\right]$

$\Delta T=-0.005T$

Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

A ring whose diameter is 1 meter, oscillates simple harmonically in a vertical plane about a nail fixed at its circumference and perpendicular to plane of ring. The time period will be

  1. 1/4 sec

  2. 1/2 sec

  3. 2sec

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a physical pendulum (ring oscillating about a nail on its circumference), the time period T = 2*pi * sqrt(I / mgd). Here, I = I_cm + md^2 = (1/2)mr^2 + mr^2 = (3/2)mr^2. With d = r, T = 2*pi * sqrt((3/2)mr^2 / mgr) = 2*pi * sqrt(3r / 2g). With diameter 1m, r = 0.5m. T = 2*pi * sqrt(1.5 / 9.8) approx 2 seconds.

Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

Three similar oscillators, A, B, C have the same small damping constant $r$, but different natural frequencies $\omega _0 = (k/m)^{\frac{1}{2}} : 1200 Hz, 1800 Hz, 2400 Hz$. If all three are driven by the same source at $1800 Hz$, which statement is correct for the phases of the velocities of the three?

  1. $\phi _A = \phi _B = \phi _c$
  2. $\phi _A < \phi _B = 0 < \phi _c$
  3. $\phi _A > \phi _B = 0 > \phi _c$
  4. $\phi _A > \phi _B > 0 > \phi _c$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$tan \phi = \dfrac{\omega L- \dfrac{1}{\omega C}}{R}$

If $\omega < \omega _0 \Longrightarrow$ circuit is capacitive

$\Longrightarrow$ it leads voltage
$\Longrightarrow$ velocity leads force.
if $\omega = \omega _0 \phi = 0$
if $\omega > {\omega} _0$ velocity lags behind force.

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The time taken to complete $20$ oscillations by a seconds pendulum is: 

  1. $20s$
  2. $50s$
  3. $40s$
  4. $5s$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that the time period of a seconds pendulum is $T=2$ sec. One second for a swing in one direction and one second for the return swing. 

Thus, time taken to complete one oscillation is $2$ sec.
Hence, time taken to complete 20 oscillations is $2\times 20=40$ sec.

Multiple choice physics measurements and experimentation a few applications of linear shm simple pendulum example of simple harmonic motion

The length of a second's pendulum on the surface of the earth is equal to 99.49 cm. True or false.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The time period of seconds pendulum T = 2 seconds, acceleration due to gravity at earth g= 980 $\dfrac { cm }{ { s }^{ 2 } } $,It '$l$' is the length of pendulum,

$l=\dfrac { { T }^{ 2 }g }{ 4{ \pi  }^{ 2 } } \ \Rightarrow l=\dfrac { 4\times 980 }{ 4\times \left( \dfrac { 22 }{ 7 }  \right) ^{ 2 } } =\dfrac { 4\times 980\times 49 }{ 4\times 489 } =99.49$