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

Two dissimilar spring fixed at one end are stretched by 10cm and 20cm respectively, when masses ${ m } _{ 1 }$ and ${ m } _{ 2 }$ are suspended at their lower ends. When displaced slightly from their mean positions and released, they will oscillate with period in the ratio

  1. 1 : 2

  2. 2 : 1

  3. 1 : 1.41

  4. 1.41 :4
Reveal answer Fill a bubble to check yourself
A Correct answer
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 bob of mass  $\mathrm { M }$  is hung using a string of length  $\mathrm { l }.$  A mass  $m$  moving with a velocity  $u$  pierces through the bob and emerges out with velocity  $\dfrac { u } { 3 } ,$  The frequency of oscillation of the bob considering as amplitude  $A$ is

  1. $2 \pi \sqrt { \dfrac { 3 m u } { 2 M A } }$
  2. $\dfrac { 1 } { 2 \pi } \sqrt { \dfrac { 2 m } { 3 M A } }$
  3. $\dfrac { 1 } { 2 \pi } \left( \dfrac { 2 m u } { 3 M A } \right)$

  4. cannot be found
Reveal answer Fill a bubble to check yourself
A Correct answer
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  body of mass 0.98 Kg is suspended from a spring of spring constant K = 2N/m. Then the period is. 

  1. 4.9s

  2. 4.4s

  3. 5.2s

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The period of a mass-spring system is T = 2 * pi * sqrt(m/k). Given m = 0.98 kg and k = 2 N/m, T = 2 * pi * sqrt(0.98/2) = 2 * pi * sqrt(0.49) = 2 * pi * 0.7 = 1.4 * pi. Using pi approx 3.14, T is approx 4.396s, which rounds to 4.4s.

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

Two particles  $A$  and  $B$  of equal masses are suspended from two massless springs of spring constants  $k _ { 1 }$  and  $k _ { 2 }$  respectively. If the maximum velocities during oscillations are equal, the ratio of the amplitudes of  $A$  and  $B$  is

  1. $\sqrt { k _ { 1 } / k _ { 2 } }$
  2. $k _ { 1 } / k _ { 2 }$
  3. $\sqrt { k _ { 2 } / k _ { 1 } }$
  4. $k _ { 2 } / k _ { 1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Maximum velocity v_max = A * omega = A * sqrt(k/m). Since masses are equal, v_max is proportional to A * sqrt(k). For equal velocities, A1 * sqrt(k1) = A2 * sqrt(k2), so A1/A2 = sqrt(k2/k1).

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 of mass $4\, kg$ hangs from a spring and oscillates with a period $0.5$ second. On the removed of the body, the spring is shortened by

  1. $6.4\, cm$
  2. $6.2\, cm$
  3. $6.8\, cm$
  4. $7.1\, cm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The period T = 2 * pi * sqrt(m/k). Given T = 0.5s and m = 4kg, 0.5 = 2 * pi * sqrt(4/k). Squaring both sides: 0.25 = 4 * pi^2 * (4/k), so k = 64 * pi^2. The extension x = mg/k = (4 * 9.8) / (64 * pi^2) approx 39.2 / 631.65 approx 0.06206 m, which is 6.2 cm.

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

A mass m is suspended from the two coupled springs connected in series. The force constant for springs are $ K _1 and K _2 $. The time period of the suspended mass will be-

  1. $ T = 2 \pi \sqrt { \left( \dfrac { m }{ k _ 1-k _ 2 } \right) } $
  2. $ T = 2 \pi \sqrt { \left( \dfrac { m }{ k _ 1+k _ 2 } \right) } $
  3. $ T = 2 \pi \sqrt { \left( \dfrac { m\left( k _ 1+k _ 2 \right) }{ k _{ 1 }k _{ 2 } } \right) } $
  4. $ T = 2 \pi \sqrt { \left( \dfrac { mk _ 1k _ 2 }{ k _{ 1 }+k _{ 2 } } \right) } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For springs in series, the effective spring constant k_eff is given by 1/k_eff = 1/k1 + 1/k2, which simplifies to k_eff = (k1 * k2) / (k1 + k2). The time period T = 2 * pi * sqrt(m/k_eff) = 2 * pi * sqrt(m * (k1 + k2) / (k1 * k2)).