Tag: oscillations

Questions Related to oscillations

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

Time period of a sample pendulum is T, time taken by it to complete 3/8 oscillations (in terms of distance traveled) is 

  1. $
    \cfrac { 3 T } { 12 }
    $

  2. $
    \cfrac { 3 T } { 8 }
    $

  3. $
    \cfrac { 7 T } { 12 }
    $

  4. $
    \cfrac { 5 T } { 12 }
    $


Correct Option: D

A metallic disc oscillates about an axis through its edge in it's own plane. The equivalent length of the disc as a pendulum is

  1. $r$

  2. $\dfrac r3$

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

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


Correct Option: D

The bob of a simple pendulum executes  $S H M$  in water with a period  $t,$  while the period of oscillation of the bob is  $t _{ 0 }$  in air. Neglecting the frictional force of water and given that the density of the bob is  $( 4 / 3 ) \times 1000 kg / { m } ^ { 3 }.$  What relationship between  $t$  and  $t _ { 0 }$  is true ?

  1. $t = t _ { 0 }$

  2. $t = 4 t _ { 0 }$

  3. $t = 2 t _ { 0 }$

  4. $t = t _ { 0 } / 2$


Correct Option: C

A simple pendulum is released when $\theta = \pi/6$. The time period of oscillation is

  1. $\displaystyle 2\pi\sqrt{\frac{l}{g}}$

  2. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{293}{288}\right)$

  3. $\displaystyle 2\pi\sqrt{\frac{l}{g}}\left(\frac{288}{293}\right)$

  4. none of these


Correct Option: B
Explanation:

For large amplitudes, the time period is given by 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(1+\dfrac{\theta ^{2}}{16})$
Substitute $\theta =\dfrac{\pi }{6}$, we get answer as 
$T={2\pi }{\sqrt{\dfrac{L}{g}}}(\dfrac{293}{288})$
Option B is correct.