Tag: oscillatory motion

Questions Related to oscillatory motion

When two blocks connected by a spring move towards each other under mutual interaction:

  1. Their velocities are equal and opposite

  2. Their accelerations are equal and opposite

  3. The forces acting on them are equal and opposite

  4. Their momenta are equal and opposite.


Correct Option: C
Explanation:

If we take the two blocks plus spring as the system there is no external force acting on this system.
The accelerations will be equal and opposite if masses are equal. Since the forces are internal, they will be equal and opposite.

Two springs have their force constants ${ K } _ { 1 }$ and ${ K } _ { 2 }.$ Both are stretched till their elastic energies are equal. Then,ratio of stretching forces ${ K } _ { 1 } / { K } _ { 2 }$ is equal to:

  1. $K _ { 1 } / K _ { 2 }$

  2. $\mathbf { K } _ { 2 } : \mathbf { K } _ { 1 }$

  3. $\sqrt { K _ { 1 } } : \sqrt { K _ { 2 } }$

  4. $\mathbf { K } _ { 2 } ^ { 2 } : \mathbf { K } _ { 2 } ^ { 2 }$


Correct Option: C

One end of a light spring of force constant K is fixed to ceiling the other end is fixed to block of mass M initially the spring is relaxed the work done by the external agent to lower the Hanging body of mass M slowly till it comes to equilibrium is

  1. $3 m^2 g^2/ 2k$

  2. $m^2 g^2/ 2k$

  3. $-3 m^2 g^2/ 2k$

  4. $- m^2 g^2/ 2k$


Correct Option: A

A spring oscillates with frequency $1$ cycle per second. What approximate length must a simple pendulum have to oscillate with that same frequency?

  1. 25 cm

  2. 50 cm

  3. 67 cm

  4. 90 cm


Correct Option: B

Two identical springs are fixed at one end and masses $1$ $kg$ and $4$ $kg$ are suspended at their other ends. They are both stretched down from their mean position and let go simultaneously. If they are in the same phase after every $4$ seconds then the springs constant $k$ is 

  1. $\pi \dfrac { N }{ m } $

  2. ${ \pi }^{ 2 }\dfrac { N }{ m } $

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

  4. $given$ $data$ $is$ $insufficient$


Correct Option: C

A body is attached to the lower end of a vertical spiral spring and it is gradually lowered to its equilibrium position.This stretches the spring by a length d.If the same body attached to the same spring is allowed to fall suddenly, what would be the maximum stretching in this case?

  1. d

  2. 2d

  3. 3d

  4. 1/2d


Correct Option: B

A spring $40\ mm$ long is stretched by the application of a force. If $10\ N$ force required to stretch the spring through $1\ mm$, then work done in stretching the spring through $40\ mm$ is:

  1. 84 J

  2. 68 J

  3. 23 J

  4. 8 J


Correct Option: D

A mass of $0.98kg$ suspended using a spring of constant $K=300Nm^{-1}$ is hit by a bullet of 20gm moving with a velocity $3.0m/s$ vertically. The bullet gets embedded and oscillates with the mass .  The amplitude of oscillation will be-

  1. $0.15cm$

  2. $0.12cm$

  3. $1.2cm$

  4. $12m$


Correct Option: A

A spring of force constant K is cut into two pieces such that one piece is double the length of the other Then the long piece will have a force constant of

  1. 2 k/3

  2. 3 k/2

  3. 3 k

  4. 6 k


Correct Option: B
Explanation:
Length of the spring $= L$
Force constant of spring $= K$
Ratio in which spring is cut $= 1 : 2$
Length of larger piece $= 2L / (2 + 1) = 2L/3$
Force constant of larger piece $= K’$
Force constant ∝ 1 / Length of the spring
$K / K’ = (2L / 3) / L$
$K / K’ = 2 / 3$
$K’ = 3K / 2$
$K’ = 1.5 K$
Force constant of larger piece is $1.5 K$

The potential energy of a particle executing  $S.H.M$ is $2.5 J$.

When its displacement is half of amplitude the total energy of the particle  be

  1. 18 J

  2. 15 J

  3. 10 J

  4. 12 J


Correct Option: C
Explanation:

$\begin{array}{l} We\, \, know, \ \dfrac { { potential\, \, energy\, \left( U \right)  } }{ { Total\, \, energy\left( E \right)  } } =\dfrac { { \dfrac { 1 }{ 2 } m{ \omega ^{ 2 } }{ y^{ 2 } } } }{ { \dfrac { 1 }{ 2 } m{ \omega ^{ 2 } }{ a^{ 2 } } } } =\dfrac { { { y^{ 2 } } } }{ { { a^{ 2 } } } }  \ So, \ \dfrac { { 2.5 } }{ E } =\dfrac { { { { \left( { \dfrac { a }{ 2 }  } \right)  }^{ 2 } } } }{ { { a^{ 2 } } } }  \ E=10\, \, J \end{array}$