Tag: angular simple harmonic motion

Questions Related to angular simple harmonic motion

A rod of mass 'M' and length '2L' is suspended at its middle by a wire. It exhibits torsional oscillations; If two masses each of 'm' are
attached at distance $'L/2'$ from its centre on both sides, it reduces the oscillation frequency by $20\%$. The value of ratio m/M is close to :

  1. 0.175

  2. 0.375

  3. 0.575

  4. 0.775


Correct Option: B
Explanation:

Frequency of torsonal oscillations is given by 
$f = \dfrac{k}{\sqrt{I}}$
$f _1 = \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12}}}$
$f _2 =  \dfrac{k}{\sqrt{\dfrac{M (2L)^2}{12} + 2m \left(\dfrac{L}{2} \right)^2}}$
$f _2 = 0.8 f _1$
$\dfrac{m}{M} = 0.375$

Find the time period of oscillations of a torsional pendulum, if the torsional constant of the wire is K = 10$\pi^2$J/rad. The moment of inertia of rigid body is 10 Kg m$^2$ about the axis of rotation.

  1. 2 sec

  2. 4 sec

  3. 16 sec

  4. 8 sec


Correct Option: A
Explanation:

Time period of a torsional pendulum is given by
$T = 2 \pi \sqrt{\dfrac{I}{k}}$
$\Rightarrow T=2 \pi \sqrt{\dfrac{10}{10\pi^2}}=2 sec$

A clock which has a pendulum made of brass keep correct time at ${30^0}C$? How many seconds it will gain or lose in day if the temperature falls to ${0^0}C$.

  1. It will lose $23.32$ sec per day

  2. It will gain $23.32$ sec per day

  3. It will lose $50$ sec per day

  4. It will gain $50$ sec per day


Correct Option: A

A simple pendulum of length $1$m has a bob of mass $100$g. It is displaced through an angle of $60^o$ from the vertical and then released . Find out K.E. of bob when it passes through mean position.

  1. $0.12$J

  2. $0.24$J

  3. $0.36$J

  4. $0.55$J


Correct Option: D
Explanation:

Length of simple pendulum $=1m$

Mass $=1w\;gm=0.1kg$
It is displaced through as angle of $60$ for vertical 
Height of pendulum at starting position $=$ length $(1-ws\;60)$
                                                                   $=1\left( 1-0.5\right)$
                                                                   $=0.5m$
Potential energy $=mgh=0.1\times 10\times 0.5 =0.55$
when it is released and it reaches mean position its potential energy at starting point is converted to kinetic energy.
so K.E. of bob at mean position $=0.55.$
Hence, the answer is $0.55.$

A pendulum is formed by pivoting a long thin rod of length L and mass m about a point P on the rod which is a distance d above the center of the rod as shown. 
Now answer the following questions. 
1. The time period of this pendulum when d = L/2 will be

  1. $2\pi\sqrt { \dfrac { 2\ell }{ 3g }}$

  2. $2\pi\sqrt { \dfrac { 3\ell }{ 2g }}$

  3. $4\pi\sqrt { \dfrac { \ell }{ 3g }}$

  4. $\dfrac {2\pi} {3} \sqrt { \dfrac { 2\ell }{ g }}$


Correct Option: C

A uniform rope of mass M =0.1 kg and length L=10 m hangs from the ceiling. [$  g=10 m /{ s }^{ 2 } $]

  1. Speed of the transverse wave in the rope increases linearly from bottom to the top with distance.

  2. Speed of the transverse wave in the rope decreases linearly from bottom to the top with distance.

  3. Speed of the transverse wave in the rope remains constant along the length of the rope

  4. Time taken by the transverse wave to travel the full length of the rope is 2 sec


Correct Option: D

The time period of a simple pendulum is 2.5 second. What will be it's total number of oscillations in 50 seconds ?

  1. 10

  2. 15

  3. 20

  4. None of these.


Correct Option: C

A simple pending of length l has a bob of mass m, with a charge q on it . A  vertical sheet of charge, with surface charge density $\sigma $ passes string makes an angle $\theta $ with the vertical , then 

  1. $\quad tan\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $

  2. $\quad tan\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $

  3. $\quad cot\theta =\dfrac { \sigma q }{ 2{ \epsilon } _{ 0 }mg } $

  4. $\quad cot\theta =\dfrac { \sigma q }{ { \epsilon } _{ 0 }mg } $


Correct Option: A

The bob cf a simple pendulum is a spherical hollowe bal filled with water A pluyged hole near the bouthmol th oscilloting bob gets suddenly unplugged. During observation, till water is coming out, the time period of would 

  1. First increase and then decrease to the original value

  2. first decrease and then increase to the original value

  3. remain unchanged

  4. none of these


Correct Option: A

If the simple pendulum maximum kinetic energy of length 'I'  has maximum angular displacement $\theta $ then the maximum kinetic energy of the bob of mass 'm' is:

  1. $m g / ( 1 - \cos \theta )$

  2. $\operatorname { mg } l ( 1 - \sin \theta )$

  3. $m g / ( 1 + \cos \theta )$

  4. $m g l ( 1 + \sin \theta )$


Correct Option: A