Tag: motion of a mass suspended by two springs

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Questions Related to motion of a mass suspended by two springs

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $1.5 K$
Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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. $\pi \sqrt{\cfrac{m}{k}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In free fall, the spring and mass are in the same frame of reference. The spring is not stretched by gravity, so the oscillation frequency and period remain the same as in a gravity-free environment.

Multiple choice physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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. ${ T } _{ 2 }=\sqrt { 2 } { T } _{ 1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Parallel: Kp = k + k = 2k. T1 = 2 * pi * sqrt(m/2k). Series: Ks = (k*k)/(k+k) = k/2. T2 = 2 * pi * sqrt(m/(k/2)) = 2 * pi * sqrt(2m/k). T2 = 2 * T1.

Multiple choice physics simple harmonic motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

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
Reveal answer Fill a bubble to check yourself
B Correct answer