Tag: example of simple harmonic motion

Questions Related to example of simple harmonic motion

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:

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

A force of 6.4 N stretches a vertical spring by 0.1 m. The mass that must be suspended from the spring so that it oscillates with a period of ($\pi/4$) sec is:  

  1. $(\pi/4)$ kg

  2. 1 kg

  3. $(1 / \pi)$

  4. 10 kg


Correct Option: B
Explanation:

$\begin{array}{l} k=\frac { f }{ x } =\frac { { 6.4 } }{ { 0.1 } } =64 \ T=2\pi \sqrt { \frac { m }{ k }  }  \ \frac { \pi  }{ 4 } =2\pi \sqrt { \frac { m }{ { 64 } }  }  \ m=1\, kg \end{array}$

Hence,
option $(B)$ is correct answer.

A spring of force constant $800 Nm^{-1}$ has an extension of 5 cm . The work done in extending it from 5 cm to 15 cm is

  1. 16 J

  2. 8 J

  3. 32 J

  4. 24 J


Correct Option: B

A spring of spring constant ($k$) is attached to a block of mass ($m$). During free fall its time period of oscillations will be

  1. Zero

  2. Infinite

  3. $2\pi \sqrt{\cfrac{m}{k}}$

  4. $\pi \sqrt{\cfrac{m}{k}}$


Correct Option: C

A man weighing 60 kg stands on the horizontal platform of a spring balance. The platform starts executing simple harmonic motion of amplitude 0.1 m and frequency $2/ \pi$ Hz. Which of the following statements is correct?

  1. The spring balance reads the weight of man as 60kg

  2. The spring balance reading fluctuates between 60 kg. and 70 kg

  3. The spring balance reading fluctuates between 50 kg and 60 kg

  4. The spring balance reading fluctuates between 50 kg and 70 kg


Correct Option: A
Explanation:

Here,

The option $A$ is the correct answer because

as you know spring balance observe normal reaction between contacting surface. it is effected only when lift accelerated or decelerated . 
if lift is moving upward with acceleration $a $
then, observation of spring balance will be $= m(g + a)$ , where m is mass of man 
when lift is moving downward with acceleration a then, observation of spring balance will be $= m(g - a) .$
but when lift is moving upward or donward with constant velocity then, observation will be remain same.
hence, observation of man's weight is $60kg$ on spring balance.

Two identical springs are attached to a mass and the system is made to oscillate. ${ T } _{ 1 }$ is the time period when springs are joined in parallel and ${ T } _{ 2 }$ is the time period when they are joined in series then

  1. ${ T } _{ 1 }=2{ T } _{ 2 }$

  2. ${ T } _{ 1 }=\sqrt { 2 } { T } _{ 2 }$

  3. ${ T } _{ 2 }=2{ T } _{ 1 }$

  4. ${ T } _{ 2 }=\sqrt { 2 } { T } _{ 1 }$


Correct Option: C

A loaded spring gun. Initially at rest on a horizontal frictioneles surface fires a marble of  mass m in at an angle of elevation ${ 0 }^{ o }$. The mass of the gun is M that of the marble is m and its muzzle velocity of the marble is ${ V } _{ 0 }$ then Velocity of the gem just after the firing is 

  1. $\dfrac { m{ v } _{ 0 } }{ M } $

  2. $\dfrac { m{ v } _{ 0 }\cos { \theta } }{ M } $

  3. $\dfrac { m{ v } _{ 0 }\cos { \theta } }{ M+m } $

  4. $\dfrac { m{ v } _{ 0 }\cos { 2\theta } }{ M+m } $


Correct Option: A

A block tied between two springs is in equilibrium. If upper spring is cut then the acceleration of the block just after cut is 6 ${ m/s }^{ 2 }$ downwards. Now, if instead of upper spring, lower spring is cut then the magnitude of acceleration of the block just after the cut will be : (Take g = 10 ${ m/s }^{ 2 }$)

  1. 16 ${ m/s }^{ 2 }$

  2. 4 ${ m/s }^{ 2 }$

  3. Cannot be determined

  4. None of these


Correct Option: B