Tag: oscillatory motion

Quizzes (7) Questions (124)

Questions Related to oscillatory motion

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

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

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

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. $\mathbf { K } _ { 2 } ^ { 2 } : \mathbf { K } _ { 2 } ^ { 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Elastic energy U = F^2 / (2k). If U1 = U2, then F1^2 / (2k1) = F2^2 / (2k2). Rearranging gives (F1/F2)^2 = k1/k2, so F1/F2 = sqrt(k1)/sqrt(k2).

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

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. $- m^2 g^2/ 2k$
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 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
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A spring with frequency f = 1 Hz has a period T = 1 s. A simple pendulum having the same frequency must have length L given by T = 2*pi*sqrt(L/g). Setting T = 1 s and g = 9.8 or pi^2 gives L = g / (4*pi^2) approximately equal to 0.25 meters, or 25 cm.

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 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. $given$ $data$ $is$ $insufficient$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The time periods are T1 = 2*pi*sqrt(1/k) and T2 = 2*pi*sqrt(4/k) = 2*T1. They are in the same phase after every 4 seconds, meaning 4 seconds is a common multiple of their periods. Solving the relations yields the spring constant k in terms of pi squared.

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

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

Gradual lowering reaches equilibrium at d = mg/k. Sudden release results in maximum extension at 2d because the potential energy lost by the mass (mg * 2d) equals the energy stored in the spring (1/2 * k * (2d)^2).

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 $40\ mm$ 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
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Force constant k = F/x = 10N / 1mm = 10,000 N/m. Work done W = 1/2 * k * x^2 = 0.5 * 10,000 * (0.04m)^2 = 5,000 * 0.0016 = 8 J.

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

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