Physics

Spring Mass Systems

172 Questions

A spring mass system is a key physics concept used to study oscillations and simple harmonic motion. It involves understanding spring constants, damping, and series or parallel combinations. These principles are frequently tested in engineering entrance examinations.

Series and parallel springsSpring constant calculationsDamped oscillationsSpring compression energyElevator systems

Spring Mass Systems Questions

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 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  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)).

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 block of mass m is suspended separately by two different spring have time period $ t _1 and t _2 $ . if same mass is connected to parallel combination of both springs , then its time period is given by

  1. $ \dfrac {t _1t _2}{t _1 +t _2} $
  2. $ \dfrac {t _1t _2}{\sqrt {t^2 _1+ t^2 _1} } $
  3. $ \sqrt {\dfrac { t _1t _2}{ t _1 +t _2}} $
  4. $\sqrt {(t _1)^2 + (t _2)^2} $
Reveal answer Fill a bubble to check yourself
D 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 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 } } $

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 long metallic wire of length $L$ area of cross-section $A$ and Young's modulus $Y$ is tied to the ceiling. The other end is tied to a massless spring of force constant $k$. A mass $m$ hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by-

  1. $\displaystyle 2\pi \sqrt{\frac{m}{k}}$
  2. $\displaystyle 2\pi \sqrt{\frac{mYA}{kL}}$
  3. $\displaystyle 2\pi \sqrt{\frac{mk}{YA}}$
  4. $\displaystyle 2\pi \sqrt{\frac{m(kL+YA)}{kYA}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$F = \dfrac{YA\Delta l}{L} = k _2 \Delta l$
we can consider the system as two springs in series hence 
$\dfrac{1}{k _{eq}} = \dfrac{1}{k _1} +\dfrac{1}{k _2}$
$=\dfrac{1}{k} + \dfrac{L}{YA} = \dfrac{YAk +kL}{YAk}$

$ T = 2\pi \sqrt{\dfrac{m}{k _{eq}}} = 2\pi \sqrt{\dfrac{m(YAk + kL)}{YAk}}$
Multiple choice physics oscillatory motion motion of a mass suspended by two springs example of simple harmonic motion oscillations due to a spring

The frequency $f$ of vibrations of a mass $m$ suspended from a spring of spring constant $k$ is given by $f = Cm^xk^y$, where $C$ is a dimensionless constant. The values of $x$ and $y$ are respectively:

  1. $\dfrac{1}{2}, \dfrac{1}{2}$
  2. $-\dfrac{1}{2}, -\dfrac{1}{2}$
  3. $\dfrac{1}{2}, -\dfrac{1}{2}$
  4. $-\dfrac{1}{2}, \dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know $F=-KK\Rightarrow dim\left( K \right) =\left[ { MLT }^{ -2 } \right] \left[ { L }^{ -1 } \right] ={ ML }^{ 0 }{ T }^{ -2 }$
$dim\left( M \right) ={ ML }^{ 0 }{ T }^{ 0 }$    $dim\left( f \right) =\left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 } \right] $
$f={ Cm }^{ x }{ K }^{ y }\Rightarrow { M }^{ 0 }{ L }^{ 0 }{ T }^{ -1 }={ \left[ { ML }^{ 0 }{ T }^{ 0 } \right]  }^{ k }{ \left[ { ML }^{ 0 }{ T }^{ -2 } \right]  }^{ y }$
$\Rightarrow$  Comparing powers of $M,L$ and $T$ gives,
$x+y=0\quad \quad -2y=-1\quad \Rightarrow \quad y=\dfrac { 1 }{ 2 } $
and $x=-1/2$