Tag: example of simple harmonic motion

Questions Related to example of simple harmonic motion

Time period of a disc about a tangent parallel to the diameter is same as the time period of a simple pendulum. The ratio of radius of disc to the length of pendulum is :

  1. $\dfrac { 1 } { 4 }$

  2. $\dfrac { 4 } { 5 }$

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

  4. $\dfrac { 1 } { 2 }$


Correct Option: C

A pendulum of mass $m$ hangs from a support fixed to a trolley. The direction of the string (i.e.., angle $\theta$) when the trolley rolls up a plane of inclination $\alpha$ with acceleration $'a'$ is

  1. Zero

  2. $\tan^{-1} \alpha$

  3. $\tan^{-1}\dfrac{a+g \sin \alpha}{g \cos \alpha}$

  4. $\tan^{-1}\dfrac{a}{g}$


Correct Option: C

The oscillations of a pendulum slow down due to

  1. the force exerted by air and friction at the support

  2. the foce exerted by air only

  3. the forces exerted by friction at the support only

  4. none of these


Correct Option: A
Explanation:

The pendulum slows down due to friction force exerted by air and friction at the support.

The pendulum of a certain clock has time period $2.04 s$. How fast or slow does the clock run during $24$ hour?

  1. $28.8$ minutes slow

  2. $28.8$ minutes fast

  3. $14.4$ minutes fast

  4. $14.4$ minutes slow


Correct Option: A

A string of simple pendulum can bear maximum tension that is twice the weight of the bob. What is the maximum angle $(\theta)$ with which it can oscillate?

  1. $0^0$

  2. $45^0$

  3. $60^0$

  4. $90^0$


Correct Option: B

A bob is suspended from an ideal string of length $l$. Now it is pulled to a side through $60^{o}$ to vertical and rotates along a horizontal circle. Then its period of revolution is

  1. $ 2\pi \sqrt{ l/g }$

  2. $ \pi \sqrt{ l/2g }$

  3. $\pi\sqrt{ 2l/g }$

  4. $\pi\sqrt{ l/g }$


Correct Option: C

Two simple pendulums have time period $4\ s$ and $5\ s$ respectively. If they started simultaneously from the mean positive in the same direction, then the phase difference between them by the time the larger one completes one osicillation is

  1. $\dfrac {\pi}{6}$

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

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

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


Correct Option: A

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)