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

Frequency of a block in spring-mass system is $\displaystyle \upsilon $, if it is taken in a lift slowly accelerating upward, then frequency will 

  1. decrease

  2. increase

  3. remain constant

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\omega=2\pi\sqrt{\dfrac{K}{M}}$

Frequency is independent of gravity
Hence it will remain constant

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

A uniform spring has certain mass suspended from it and it's period of vertical oscillations is ${t} _{1}$. The spring is now cut in $2$ parts having lengths in ratio $1:2$  and these springs are now connected in series and then in parallel. find out the ratio of the time period of these two ossillation?

  1. $1$
  2. $\sin \theta$
  3. $\sqrt {\dfrac {2}{9}}$
  4. $\sqrt {\dfrac {9}{2}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let $k$ be initial force constant of spring,${k} _{1}$ and ${k} _{2}$ be the force constant of neew springs
We can derive,
$ kl= constant $
$\Rightarrow \dfrac{{x} _{1}}{{x} _{2}}=1/2$
$\Rightarrow \dfrac{{k} _{1}}{{k} _{2}}=2$     ........$(1)$
$so k _1=3k, k _2=3k/2 $
As initially these lengths were in series:
$\dfrac{1}{k}=\dfrac{1}{{k} _{1}}+\dfrac{1}{{k} _{2}}$
$\Rightarrow \dfrac{1}{k' _1}=\dfrac{1}{3{k} _{1}}+\dfrac{2}{3{k} _{1}}$
$\Rightarrow \dfrac{1}{k' _1}=\dfrac{1}{{k} _{1}}$
$\Rightarrow {k'} _{1}= k\ $
When these two stringd are connected in parallel,
${k _2}^{\prime}={k} _{1}+{k} _{2}$

${k _2}^{\prime}=\dfrac{3k}{2}+3k$

${k _2}^{\prime}=\dfrac{9k}{2}$

Time period is 
$\dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{k' _1}{{k' _2}^{\prime}}}$
$\Rightarrow \dfrac{{T _1}^{\prime}}{T' _2}=\sqrt{\dfrac{2}{9}}$
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 $1.5$ kg block at rest on a tabletop is attached to a horizontal spring having a spring constant of $19.6$ N/m. The spring is initially unstretched. A constant $20.0$ N horizontal force is applied to the object causing the spring to stretch.Determine the speed of the block after it has moved $0.30$ m from equilibrium if the surface between the block and the tabletop is frictionless.

  1. $2.61\ m/s$
  2. $3.61\ m/s$
  3. $7.61\ m/s$
  4. $8.1\ m/s$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This system will exhibits S.H.M with angular frequency $\omega =\sqrt { \cfrac { k }{ m }  } =\sqrt { \cfrac { 19.6 }{ 1.5 }  } $

$k$= spring constant 
$m$= mass of the body
The string stretched by the maximum A restoring force at maximum stretch= force acting on body
$\Rightarrow kA=F$
$\Rightarrow A=\cfrac { F }{ k } =\cfrac { 20 }{ 19.6 } $
$F=20N$
$K=19.6\quad N/m$
Now if x is displacement from mean position, the velocity is given by:
$v=\omega =\sqrt { ({ A }^{ 2 }-{ x }^{ 2 }) } $
$v=\omega =\sqrt { \cfrac { 19.6 }{ 1.5 } ({ \cfrac { 20 }{ 19.6 }  }^{ 2 }-{ 0.3 }^{ 2 }) } $
$=3.523m/s$

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

An infinite number of springs having force constants as K, 2K, 4K, 8K, .......$\displaystyle \infty $ respectively are connected in series; then equivalent spring constant is 

  1. K

  2. 2K
  3. $\displaystyle \frac{K}{2}$
  4. $\displaystyle \infty $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For the springs connected in series

$\dfrac{1}{K _{eq}}=\dfrac{1}{K}+\dfrac{1}{2K}+\dfrac{1}{4K}+\dfrac{1}{8K}+......$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(1+\dfrac{1}{2}+\dfrac{1}{4}+.....)$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(\dfrac{1}{1-\dfrac{1}{2}})$
$\dfrac{1}{K _{eq}}=\dfrac{1}{K}(2)$
$K _{eq}=\dfrac{K}{2}$

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 $m$ is suspended from a spring of spring constant $k$. A damping force proportional to the velocity exerts itself on the mass. An appropriate representation of the motion is 

  1. $ m \dfrac{d^2x}{dt^2} - c \dfrac{dx}{dt} + kx = 0$
  2. $ m \dfrac{d^2x}{dt^2} + c \dfrac{dx}{dt} - kx = 0$
  3. $ m \dfrac{d^2x}{dt^2} - c \dfrac{dx}{dt} - kx = 0$
  4. $ m \dfrac{d^2x}{dt^2} + c \dfrac{dx}{dt} + kx = 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The forces on the mass $m$ are due to spring and damper.

Suppose the mass moves by a distance $x$ from equilibrium and travels with a velocity $\displaystyle \frac{dx}{dt}$, then

Force due to spring is $-kx$ and force due to damper is $\displaystyle -c\frac{dx}{dt}$

By Newton's Second Law of motion, we have $m\displaystyle \frac{d^2x}{dt^2} = -kx-c\frac{dx}{dt}$

Thus, the equation of motion is $\displaystyle m\frac{d^2x}{dt^2}+c\frac{dx}{dt}+kx=0$
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 $m$ attached to the spring experiences a drag force proportional to its velocity and an external force $F(t) = F _o \cos \omega _ot$. The position of the mass at any point in time can be given by:

  1. $x(t) = c _1 \sin (\omega t + \phi) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$
  2. $x(t) = c _1 \cos (\omega t + \phi) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$
  3. $x(t) = c _1 \sin (\omega t) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$
  4. $x(t) = c _1 \cos (\omega t) + (\dfrac{F _o}{\omega ^2 - \omega _o ^2}) \cos \omega _o t$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

Initially, we already know that the displacement at any moment (Instant) of time for spring mass system is:

$x=a\sin { (\omega t+\phi ) } $ or,
$x=a\cos { (\omega t+\phi ) } $
And here an external force is experienced thus, all the four options give position at any time.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

The natural angular frequency of a particle of mass 'm' attached to an ideal spring of force constant 'K' is

  1. $\sqrt{\frac{K}{m}}$
  2. $\sqrt{\frac{m}{K}}$
  3. $\left ( \frac{K}{m} \right )^{2}$
  4. $\left ( \frac{m}{K} \right )^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Suppose you displace the particle by a distance $'x'$
The spring now exerts a force,
This provides nccenary force for $SHM$
$\Rightarrow \ F=mwe^2x=k2$ ($w:$ natural angular frequency )
$\Rightarrow \ w=\sqrt {K/m}$

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A block of mass m is hanging vertically by spring of spring constant k. If the mass is made to oscillate vertically, its total energy is:

  1. maximum at the extreme position

  2. maximum at the mean position

  3. minimum at the mean position

  4. same at all positions
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The block executes SHM. In SHM, the total energy remains constant at all positions.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A block of mass m is suspended separately by two different springs have time period ${ t } _{ 1 }$ and ${ t } _{ 2 }$. If same mass is connected to parallel combination of both springs, then its time period is $T$. Then

  1. $T = t _1 + t _2$
  2. $T^2 = t _1^2 + t _2^2$
  3. $T^{-1} = t _1^{-1} + t _2^{-1}$
  4. $T^{-2} = t _1^{-2} + t _2^{-2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the spring constant be $k _1$ and $k _2$ 
now, formula of time period of spring is given by 
$T=2\pi \sqrt{m/k}$
now, $t _{1}= 2\pi \sqrt{ m/k _{1}}$
$t _{1}^{2} = \dfrac{4 \pi^{2} m}{k _{1}} \Rightarrow k _{1}=4 \pi^{2} m/t _{1}^{2} $----$(2)$
$t _{2}= 2 \pi \sqrt{m/ k _{}}$
$t _{2}^{2} = \dfrac{4 \pi^{2} m}{k _{2}} \Rightarrow k _{2}= 4 \pi^{2} m/t _{2}^{2}$---- $(2)$
now, both are connected in parallel then ;
$k _{1}+ k _{2} \Rightarrow \dfrac{4 \pi^{2} }{t _{1}^{2}} + \dfrac{4 \pi^{2} m }{t _{2}^{2}}$
$4 \pi^{2} m \left( \dfrac{1}{t _{1}^{2}} + \dfrac{1}{t _{2}^{2}} \right)$
now, time period $=\sqrt{\dfrac{t _{1}^{2}. t _{2}^{2}}{t _{1}^{2}+ t _{2}^{2}}}$
Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Springs of spring constants K, 2K, 4K, 8K, 2048 K are connected in series. A mass 'm' is attached to one end the system is allowed to oscillation. The time period is approximately :

  1. $2\pi \sqrt{\dfrac{m}{2K}}$
  2. $2\pi \sqrt{\dfrac{2m}{2K}}$
  3. $2\pi \sqrt{\dfrac{2m}{K}}$

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
In series the equation spring constant is given by:-
$\dfrac {1}{k'}=\displaystyle \sum \dfrac {1}{k'}\Rightarrow \dfrac {1}{k'}+\dfrac {1}{k}+\dfrac {1}{2k}+\dfrac {1}{4k}+\dfrac {1}{8k}+\dfrac {1}{2048k}$
$\Rightarrow \ \dfrac {1}{k'}=\dfrac {3841}{2048k}$
or approximately $k'\simeq \dfrac {2048k}{3841}\simeq \dfrac {k}{2}$
$\therefore \ $ New time period, $T'=2\pi \sqrt {\dfrac {m}{k'}}=2\pi \sqrt {\dfrac {2m}{k}}$


Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

A body of mass $m$ has time period $T _1$ with one spring and has time period $T _2$ with another spring. if both the spring are connected in parallel and same mass is used, then new time period $T$ is given as

  1. $T^{2}= T _{1}^{2}+ T _{2}^{2}$
  2. $T= T _{1}+ T _{2}$
  3. $\dfrac{1}{T}=\dfrac{1}{ T _{1}}+\dfrac{1}{ T _{2}}$
  4. $\dfrac{1}{ T^{2}}=\dfrac{1}{ T _{1}^{2}}+\dfrac{1}{ T _{2}^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Angular frequency, $\omega =\sqrt{\dfrac{k}{m}}\ \ \ \Rightarrow \ \dfrac{2\pi }{T}=\sqrt{\dfrac{k}{m}}$

$\Rightarrow \ k=m{{\left( \dfrac{2\pi }{T} \right)}^{2}}$  where, $T$ is time period.

Net Spring constant when two spring is connected in parallel.

$ k={{k} _{1}}+{{k} _{2}} $

$\Rightarrow m{{\left( \dfrac{2\pi }{T} \right)}^{2}}=m{{\left( \dfrac{2\pi }{{{T} _{1}}} \right)}^{2}}+m{{\left( \dfrac{2\pi }{{{T} _{2}}} \right)}^{2}}$

$ \Rightarrow \dfrac{1}{{{T}^{2}}}=\dfrac{1}{{{T} _{1}}^{2}}+\dfrac{1}{{{T} _{2}}^{2}} $ 

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

When two blocks A and B coupled by a spring on a frictionless table are stretched and then released, then

  1. kinetic energy of body at any instatn after releasing is inversely proportional to their masses

  2. kinetic energy of body at any instant may or may not be inversely proportional to their masses
  3. $\cfrac { K.E\quad of\quad B }{ K.E\quad of\quad A } =\cfrac { mass\quad of\quad B }{ mass\quad of\quad A } $

  4. both (b) and (c) are correct
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Force on each block =kx

Let acceleration at any block of mass m is ,
$ma=kx$
$a=\dfrac{kx}{m}$
$\omega=\sqrt{\dfrac{k}{m}}$
velocity $v=at=\dfrac{kx}{m}t$
$KE=\dfrac{1}{2}mv^2$
$=\dfrac{1}{2}m \dfrac{k^2x^2t^2}{m^2}=\dfrac{1}{2}\dfrac{k^2x^2t^2}{m}$
$KE \alpha \dfrac{1}{m}$

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two blocks of masses $m _1$ and $m _2$ are connected by a massless spring and placed on smooth surface. The spring initially stretched and released. Then:

  1. The momentum of each particle remains constant seperately

  2. The magnitude of momentums of each body are equal to each other

  3. The mechanical energy of system remains constant

  4. Both (2) &(3)
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In a system of two blocks connected by a spring on a smooth surface, the internal spring force acts equally on both, meaning the magnitude of momentum change is equal for both. Since the system is isolated, the total mechanical energy is conserved.

Multiple choice horizontal oscillations of a mass attached to a spring oscillations due to a spring simple harmonic motion oscillations physics

Two masses $m _{1}=1\ kg$ and $m _{2}=0.5\ kg$ are suspended together by a massless spring of spring constant $12.5\ Nm^{-1}$. When masses are in  equilibrium $m _{1}$ is removed without disturbing the system. New amplitude of oscillation will be 

  1. $30\ cm$
  2. $50\ cm$
  3. $80\ cm$
  4. $60\ cm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
we have formula,

$x _2=\dfrac{m _1g}{k}$

where,

$x _2=amplitude$

$k=spring-constant$

$x _2=\dfrac{1 \times 9.8}{12.5}=0.784\approx 0.8m$

$\therefore x _2=80cm$ is the new amplitude of oscillation.