Tag: oscillations

Quizzes (16) Questions (206)

Questions Related to oscillations

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

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 

  1. zero degree
  2. $30^o$
  3. $45^o$
  4. $60^o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an accelerating frame, the equilibrium position of the pendulum shifts by an angle alpha = arctan(a/g). Given a = g/sqrt(3), alpha = arctan(1/sqrt(3)) = 30 degrees. The tension is maximum at the lowest point of the oscillation relative to the effective gravity, which is the equilibrium position (30 degrees from the vertical). However, the question asks for the angle with the vertical, and if the equilibrium is already at 30 degrees, the maximum tension occurs at that equilibrium point.

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

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

  1. 4:1

  2. 1:4

  3. 1:1

  4. 2:1
Reveal answer Fill a bubble to check yourself
D Correct answer
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 oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

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:

  1. $\sqrt{2}\pi$
  2. $16\sqrt{2} \pi$
  3. $ \dfrac{32\sqrt{2}}{\pi}$
  4. $ \dfrac{4}{\pi}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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 :

  1. $\dfrac{2\pi}{3}$
  2. $\dfrac{3}{2\pi}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{3}{\pi}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

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. $[K = mw^{2}]$.

  1. $0.02\ N/m$.
  2. $0.032\ N/m$.
  3. $0.132\ N/m$.
  4. $0.232\ N/m$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is d^2x/dt^2 + 16x = 0. Comparing this to d^2x/dt^2 + w^2x = 0, we get w^2 = 16, so w = 4 rad/s. Given mass m = 2g = 0.002 kg, the force constant K = m * w^2 = 0.002 * 16 = 0.032 N/m.