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

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 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 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

A light spring of length 20 cm and force constant 2 N/cm is placed vertically on a table. A small block of mass 1 kg falls on it. The length h from the surface of the table at which the block will have the maximum velocity is  

  1. 20 cm

  2. 15 cm

  3. 10 cm

  4. 5 cm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Maximum velocity of the block occurs at the equilibrium position where the net force is zero. Setting the downward force mg equal to the upward spring force k*x gives 1*9.8 = (2 N/cm)*x, leading to the compression at equilibrium. Subtracting this compression from the initial spring length yields the height h = 15 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 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
C Correct answer
Explanation

The time period of a spring mass system is given by T = 2 * pi * sqrt(m / k). The stretch of a spring is delta x = mg / k, which means m / k = delta x / g. Substituting this into the time period formula yields T = 2 * pi * sqrt(delta x / g), so the ratio of periods is equal to the square root of the ratio of stretches. The ratio of stretches is 10cm to 20cm, or 1 to 2, and the square root of 1/2 is approximately 1 to 1.41.

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)).

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

Two massless springs of force constants ${ k } _{ 1 }$ and ${ k } _{ 2 }$ are joined end to end. The resultant force constant $k$ of the system is

  1. $k=\dfrac { { k } _{ 1 }+{ k } _{ 2 } }{ { k } _{ 1 }{ k } _{ 2 } } $
  2. $k=\dfrac { { k } _{ 1 }-{ k } _{ 2 } }{ { k } _{ 1 }{ k } _{ 2 } } $
  3. $k=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }+{ k } _{ 2 } } $
  4. $k=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }-{ k } _{ 2 } } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In series, resultant force constant is given as
  $\dfrac { 1 }{ { k } _{ eq } } =\dfrac { 1 }{ { k } _{ 1 } } +\dfrac { 1 }{ { k } _{ 2 } } $
$\Rightarrow { k } _{ eq }=\dfrac { { k } _{ 1 }{ k } _{ 2 } }{ { k } _{ 1 }+{ k } _{ 2 } } $