Tag: oscillations

Questions Related to oscillations

The string of a simple pendulum in attached with the ceiling of a car moving on a straight horizontal raod with an acceleration $a=\dfrac {g}{\sqrt3}$, where $g$ is acceleration due to gravity near earth surface. The pendulum is made to oscillate at an angular amplitude of $30^o$. If the tension in the string is maximum when the string makes an angle $\theta$ with the vertical, then value of $\theta$ is 

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

  1. zero degree

  2. $30^o$

  3. $45^o$

  4. $60^o$

Show answer
Correct Option: A

A disc of masses m and radius 2r are suspended through a fine wire of torsional constant K. The wire is attached to the centre of the plane of the disc and given torsional oscillations. If the disc is replaced by another disc of mass 4m and radius 2r, the ratio of the time period of oscillations are

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

  1. 4:1

  2. 1:4

  3. 1:1

  4. 2:1

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Correct Option: D

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) $

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

  1. $3$ $ m s^{-1} $

  2. $
    1 / 3 \mathrm{ms}^{-1}
    $

  3. $
    1 / 2 m s^{-1}
    $

  4. $
    2 m s^{-1}
    $

Show answer
Correct Option: D
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.

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

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

  1. A simple pendulum

  2. A physical pendulum

  3. A torsional pendulum

  4. A spring-mass system

Show answer
Correct Option: A,B
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 $T = 2\pi\sqrt{m/k}$

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:

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

  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. $T = 2 \pi \sqrt{\dfrac{L}{\sqrt{g^2}- \dfrac{v^4}{R^2}}}$

Show answer
Correct Option: B
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.


The force which tries to bring a body back to its mean position is called :

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. deforming force

  2. restoring force

  3. gravitational force

  4. buoyant force

Show answer
Correct Option: B
Explanation:

At the extreme position, kinetic energy is zero and we know to throw out S.H.M.

K.E+P.E = constant
So, P.E. should be maximum at the extreme position.

A particle executes $SHM$ with a time period of $16\ s$. At time $t=2\ s$, the particle crosses the mean position while at $t=4s$, its velocity is $4ms^{-1}$. The amplitude of motion in meter is:

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $\sqrt{2}\pi$

  2. $16\sqrt{2} \pi$

  3. $ \dfrac{32\sqrt{2}}{\pi}$

  4. $ \dfrac{4}{\pi}$

Show answer
Correct Option: C
Explanation:

Let the equation of $S.H.M$ is:-


$x=a\sin\left(\dfrac{2\pi }{T}t+\phi\right)$

when $t=2s, x=0$ and $T=16s$ So,

$0=a\sin \left(\dfrac{\pi}{4}+\phi\right)$

Or $\phi=-\dfrac{\pi}{4}$

Therefore the eqn of $S.H.M$ is:-

$x=a\sin =\left(\dfrac{2\pi}{T}t-\dfrac{\pi}{4}\right)$

Now at time $t=4s, V=4m/s$

 So
$V=d\times dt=a\times \dfrac{2\pi}{T}\cos\left(\dfrac{2\pi}{T}t-\dfrac{\pi}{4}\right)$

So, $4=a\times \dfrac{2\pi}{16}\cos\left(\dfrac{\pi}{2}-\dfrac{\pi}{4}\right)$

Or, $4=a\times \dfrac{\pi}{8}\times \dfrac{1}{\sqrt{2}}$

Or $a=\dfrac{32\sqrt{2}}{\pi}$

Hence option $C$ is correct

The particle is executing S.H.M. on a line 4 cms long. If its velocity at its mean position is 12 cm/sec, its frequency in Hertz will be :

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $\dfrac{2\pi}{3}$

  2. $\dfrac{3}{2\pi}$

  3. $\dfrac{\pi}{3}$

  4. $\dfrac{3}{\pi}$

Show answer
Correct Option: B
Explanation:

Given,


$A=4cm$


$v=12cm/s$ at $x=0$ mean position

The velocity of particle performing S.H.M is given by

$v=\omega \sqrt{A^2-x^2}$

$12=\omega \sqrt{4^2-0}$

$12=4\omega$

$\omega =2\pi f=3$

$f=\dfrac{3}{2\pi}$

The correct option is B.

The different equation for linear SHM of a partial of mass $2g$ is $\dfrac {d^{2}x}{dt^{2}} + 16x = 0$. Find the force constant. $[K = mw^{2}]$.

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. $0.02\ N/m$.

  2. $0.032\ N/m$.

  3. $0.132\ N/m$.

  4. $0.232\ N/m$.

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Correct Option: B

The graph between restoring force and time in case of SHM is a

physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

  1. parabola

  2. sine curve

  3. straight line

  4. circle

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Correct Option: B
Explanation:

We know that for SHM, $x=A\sin(\omega t + \theta)$ and $F=kx=kA\sin(\omega t + \theta)$
Thus it's a sine curve.