Tag: angular simple harmonic motion

Questions Related to angular simple harmonic motion

Write the torque equation for the bob of a pendulum if it makes an angle of $\theta$ with the vertical and I is the moment of inertia of the bob w.r.t the point of suspension

  1. $I \dfrac{d^2 \theta}{dt^2}=mgL \cos \theta$

  2. $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$

  3. $I \dfrac{d^2 \theta}{dt^2}=mgL \tan \theta$

  4. $I \dfrac{d^2 \theta}{dt^2}=mg \sin \theta$


Correct Option: B
Explanation:

Taking the torque about the point of suspension, we can write $I \dfrac{d^2 \theta}{dt^2}=mgL \sin \theta$

The correct option is (b)

One end of spring of spring constant k is attached to the centre of a disc of mass m and radius R and the other end of the spring connected to a rigid wall. A string is wrapped on the disc and the end A of the string is pulled through a distance a and then released.
The disc is placed on a horizontal rough surface and there is no slipping at any contact point What is the amplitude of the oscillation of the centre of the disc?

  1. a

  2. 2a

  3. a/2

  4. none of these.


Correct Option: C
Explanation:

Displacement of the topmost point of the disc = a.
Disc undergoes rolling without slipping.
Hence the displacement of the centre of the disc = a/2
Thus the amplitude of the oscillation of the centre of the disc = a/2
Hence (C) is correct.

The angular frequency of a torsional pendulum is $\omega$ rad/s. If the moment of inertia of the object is I, the torsional constant of the wire is related to the rotational kinetic energy of the disc, if the disc was rotating with an angular velocity $\omega$ is

  1. k= 2 KE

  2. k= KE

  3. k= 4 KE

  4. k= KE/2


Correct Option: A
Explanation:

we know that $T=2 \pi \sqrt{I/k}$. Substituting the values given, we get, $k= I \omega^2 =2 \times $ kinetic energy

The correct option is (a)

A small sphere is suspended by a string from the ceiling of a car. If the car begins to move with a constant acceleration $a$, the inclination of the string with the vertical is:-

  1. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ in the direction of motion

  2. ${\tan ^{ - 1}}\left( {\dfrac{1}{2}} \right)$ opposite to the direction of motion

  3. ${\tan ^{ - 1}}\left( 2 \right)$ in the direction of motion

  4. ${\tan ^{ - 1}}\left( 2 \right)$ opposite to the direction of motion


Correct Option: B

A string of a simple pendulum can bear maximum tension that is $1.5$ times the weight of the bob. What is the maximum angle with which it can oscillate?

  1. $\cos^{-1} (0)$

  2. $\cos^{-1} (0.75)$

  3. $\cos^{-1} (0.25)$

  4. $\cos^{-1} (\sqrt {3}/2)$


Correct Option: A

A simple pendulum with a metal bob has a time period $T$. Now the bob is immersed in a liquid which is non viscous. This time the time period is $4T$. The the ratio of densities of metal bpob and that of the liquid is

  1. $15:16$

  2. $16:15$

  3. $1:16$

  4. $16:1$


Correct Option: C

A pendulum clock keeping correct time is taken to high altitudes,

  1. it will keep correct time

  2. its length should be increased to keep correct time

  3. its length should be decreased to keep correct time

  4. it cannot keep correct time even if the length is


Correct Option: C

Restoring force on the bob of a simple pendulum of mass $100\ gm$ when its amplitude is ${ 1 }^{ 0 } $ is 

  1. $0.017\ N$

  2. $1.7\ N$

  3. $0.17\ N$

  4. $0.034\ N$


Correct Option: A

Simple pendulum of large length is made equal to the radius of earth. Its period of oscillation will be then?

  1. 83.5 minutes

  2. 59.8 minutes

  3. 42.3 minutes

    1. 15 minutes

Correct Option: A
Explanation:

The time period of simple pendulum is given by:

$T=2\pi \sqrt{\dfrac{L}{g}}$

When length of pendulum is equal to the radius of earth. $R=L=6371\,\,Km=6371\times {{10}^{3}}\,Km$

So, time is

$ T=2\pi \sqrt{\dfrac{6371\times {{10}^{3}}}{10}} $

$ T=2\pi \times 798.18 $

$ T=5012.604\,seconds $

$\therefore$ $ T=83.54\,minutes $

If the length of a clock pendulum increase by $0.2\%$ due to atmospheric temperature rise, then the loss in time of clock per day is 

  1. $86.4$s

  2. $43.2$s

  3. $72.5$s

  4. $32.5$s


Correct Option: A
Explanation:
$T=2\pi \sqrt{\dfrac{l}{g}}$

$\dfrac{\Delta T}{T}\times 100 = \dfrac{1}{2}\dfrac{\Delta l}{l}\times 100 $

$\dfrac{\Delta T}{24 \times 3600}\times 100 = \dfrac{1}{2}\dfrac{0.2}{100}\times 100 $

$\Delta T=86.4s$