Tag: angular simple harmonic motion

Questions Related to angular simple harmonic motion

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 } } $


Correct Option: A

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. $13.6\ cm/s$


Correct Option: C

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


Correct Option: C

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. $\displaystyle 4: rad/s^{2}$


Correct Option: C

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. $60^o$


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

  1. 4:1

  2. 1:4

  3. 1:1

  4. 2:1


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) $

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

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

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

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


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?

  1. A simple pendulum

  2. A physical pendulum

  3. A torsional pendulum

  4. A spring-mass system


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:

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


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.