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

When a Spring of constant K  is cut into 2 equal parts then new spring constant of both the parts would be:

  1. K

  2. 2K

  3. 4K

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

$F=KL\ K=\cfrac { F }{ L } \ K\propto \cfrac { 1 }{ L } $

So when it is cut into two equal parts its length decreases to half & simultaneously spring constant increases to $2K$

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

Two identical particles each of mass $0.5\ kg$ are interconnected by a light spring of stiffness $100\ N/m,$ time period of small oscillation is

  1. $\dfrac { \pi } { 5 \sqrt { 2 } } s$
  2. $\dfrac { \pi } { 10 \sqrt { 2 } } s$
  3. $\dfrac { \pi } { 5 } s$
  4. $\dfrac { \pi } { 10 } s$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know$:$ 

$\mu  = \dfrac{{{m _1}{m _2}}}{{{m _1} + {m _2}}} = \dfrac{m}{2}$
Now$,$ $T = 2\pi \sqrt {\dfrac{\mu }{k}} $
$T = 2\pi \sqrt {\dfrac{{0.5}}{{2 \times 100}}} $
$ = \dfrac{{2\pi }}{{20}}$
$ = \dfrac{\pi }{{10}}s$
Hence,
option $(D)$ is correct answer..

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

A $100  g$ mass stretches a particular spring by $9.8\ cm,$ when suspended vertically from it. How large a mass must be attached to the spring if the period of vibration is to be $6.28\ s$?

  1. $1000\ g$
  2. ${10^5 }\ g$
  3. ${10^7}\ g$
  4. ${10^4}\ g$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\begin{array}{l} m=0.1\, \, kg,x=9.8\times { 10^{ -2 } }\, \, m,T=6.28\, \, s \ K=\dfrac { { mg } }{ x } \Rightarrow k=10 \ T=2\pi \sqrt { \dfrac { M }{ K }  } \Rightarrow 6.28=2\times 3.14\sqrt { \dfrac { M }{ { 10 } }  }  \ 1=\dfrac { M }{ { 10 } } \Rightarrow M=10\, \, kg={ 10^{ 4 } }g \end{array}$

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

Two spring-mass systems support equal mass and have spring constants $\displaystyle K _{1}$ and $\displaystyle K _{2}$. If the maximum velocities in two systems are equal then ratio of amplitude of 1st to that of 2nd is 

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

Maximum velocity $V _{max}=\omega A$

$\omega=\sqrt{\frac{K}{m}}$
$V _{1}=\sqrt{\dfrac{K _{1}}{m}}A _{1}$
$V _{2}=\sqrt{\dfrac{K _{2}}{m}}A _{2}$
It is given that both have same maximum velocity and same mass
$V _{1}=V _{2}$
$\sqrt{\dfrac{K _{1}}{m}}A _{1}=\sqrt{\dfrac{K _{2}}{m}}A _{2}$
$\dfrac{A _{1}}{A _{2}}=\sqrt{\dfrac{K _{2}}{K _{1}}}$

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 $200$ g executing SHM under the influence of a spring of spring constant $k = 90 N m^{-1}$ and a damping constant $b = 40 g s^{-1}$. Time taken for its amplitude of vibrations to drop to half of its initial values (Given, In $(1/2) = -0.693)$

  1. $7$s
  2. $9$s
  3. $4$s
  4. $11$s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given data,

mass $m=200g$
Spring constant $k=90Nm^{-1}$
Damping constant $b=40gs^{-1}$
To calculate: Time taken for the amplitude of vibration to drop to half of the initial value
We know that amplitude at any time t can be given as:

 $A(t)=A _0e^{-\dfrac{bt}{2m}}$

or $T _{1/2}=\dfrac{-0.693l×2×0.2}{40×10^{−3}}=6.93s$

Time taken for its amplitude of vibrations to drop to half of its initial values is $7s$

Multiple choice laws of vibrations of stretched strings sonometer and laws of transverse vibrations vibrations of stretched strings waves physics

A stone is hung in air from a wire, which is stretched over a sonometer. The bridges of the sonometer are 40 cm apart when the wire is in unison with a tuning fork of frequency 256 Hz. When the stone is completely immersed in water, the length between the bridges is 22 cm for re-establishing unison. The specific gravity of material of stone is 

  1. $
    \sqrt {\dfrac{{\left( {40} \right)^2 }}
    {{\left( {40} \right)^2 - \left( {22} \right)^2 }}}
    $
  2. $
    \dfrac{{\left( {40} \right)^2 }}
    {{\left( {40} \right)^2 - \left( {22} \right)^2 }}
    $
  3. $
    \dfrac{{40}}
    {{40 - 22}}
    $
  4. $
    \sqrt {\dfrac{{40}}
    {{40 - 22}}}
    $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Frequency f is proportional to sqrt(T)/L. In air, f = k*sqrt(Mg)/L1. In water, f = k*sqrt((M-m)g)/L2. Equating the two gives the ratio of weights, which leads to the specific gravity formula D/(D-d) = (L1/L2)^2.

Multiple choice angular simple harmonic motion example of simple harmonic motion oscillatory motion oscillations physics

One end of spring of spring constant k is attached to the centre of a disc of mass m and radius R and the other end of the spring connected to a rigid wall. A string is wrapped on the disc and the end A of the string is pulled through a distance a and then released.
The disc is placed on a horizontal rough surface and there is no slipping at any contact point What is the amplitude of the oscillation of the centre of the disc?

  1. a

  2. 2a

  3. a/2

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

Displacement of the topmost point of the disc = a.
Disc undergoes rolling without slipping.
Hence the displacement of the centre of the disc = a/2
Thus the amplitude of the oscillation of the centre of the disc = a/2
Hence (C) is correct.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

A body suspended from a spring balance is placed in a satellite. Reading in balance is $W _1$ when the satellite moves in an orbit of radius $R$. Reading in balance is $W _2$ when the satellite moves in an or bit of radius $2R.$ Then.

  1. $W _1 = W _2$
  2. $W _1 > W _2$
  3. $W _1 < W _2$
  4. $W _1 =2 W _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since,Both the satellites are freely falling bodies.
so,$W _1=W _2$

Multiple choice physics the essence of change forms of energy and energy conservation energy for everything forms of energy

A stretched spring possesses:

  1. kinetic energy

  2. elastic potential energy

  3. electric energy

  4. magnetic energy
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Elastic potential energy is Potential energy stored as a result of deformation of an elastic object, such as the stretching of a spring. It is equal to the work done to stretch the spring, which depends upon the spring constant k as well as the distance stretched.

Multiple choice physics oscillatory motion a few applications of linear shm simple pendulum example of simple harmonic motion

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is 15 ${ cms }^{ -1 }$ and the period is 628 milli-seconds. The amplitude of the motion in centimeters is :

  1. 3.0

  2. 2.0

  3. 1.5

  4. 1.0
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given,


$T=628ms=0.628s$


$v _{max}=15cm/s=0.15m/s$

The maximum speed of the object is given by

$v _{max}=A\omega=A\dfrac{2\pi}{T}$

Amplitude, $A=\dfrac{v _{max}T}{2\pi}$

$A=\dfrac{0.15\times 0.628}{2\times 3.14}=0.015 m$

$A=1.5cm$

The correct option is C.
Multiple choice physics oscillations a few applications of linear shm simple pendulum example of simple harmonic motion

The different equation for linear SHM of a partial of mass $2g$ is $\dfrac {d^{2}x}{dt^{2}} + 16x = 0$. Find the force constant. $[K = mw^{2}]$.

  1. $0.02\ N/m$.
  2. $0.032\ N/m$.
  3. $0.132\ N/m$.
  4. $0.232\ N/m$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation is d^2x/dt^2 + 16x = 0. Comparing this to d^2x/dt^2 + w^2x = 0, we get w^2 = 16, so w = 4 rad/s. Given mass m = 2g = 0.002 kg, the force constant K = m * w^2 = 0.002 * 16 = 0.032 N/m.

Multiple choice physics study of sound sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

Fix up one end of a slinky to a hook and hold it horizontally. Vibrate the other end to and fro along the length of the spring. If you look the spring carefully , you will see some parts of the spring are pushed closer , while some other parts are pulled apart.

  1. The portions which are pushed closer are called compressions

  2. The portions which are pulled apart are called rarefactions

  3. Both A and B are correct

  4. Neither of A and B is correct
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Compressions are the regions of high density and rarefactions are regions of low density.These compressions and rarefactions result because sound is able to reflect off fixed ends and interfere with incident waves vibrates longitudinally; the longitudinal movement of air produces pressure fluctuations. 

Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A ball is placed on a compressed spring. On releasing the spring, the ball flies away. Then which of the following is true?

  1. The ball creates its own kinetic energy

  2. Elastic potential energy of the spring is destroyed and kinetic energy of the ball is created

  3. Potential energy of the ball is destroyed and kinetic energy of the spring is created

  4. Elastic potential energy of the spring gets converted into kinetic energy of the ball
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work is done on the spring to compress it. This work is stored in the spring in the form of its elastic potential energy. When the spring is released, its elastic potential energy gets transformed into the kinetic energy of the ball and the ball flies away.

Multiple choice physics sound: production of sound oscillation - amplitude, time period and frequency of oscillation time period, frequency and amplitude of sound oscillatory and periodic motion

An object attached to one end of a spring makes 20 vibrations in 10s. Its frequency is

  1. 2Hz

  2. 10s

  3. 0.05Hz

  4. 2s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Answer is A.

The frequency (f) of a wave is the number of full wave forms generated per second. This is the same as the number of repetitions per second or the number of oscillations per second.  
In this case, an object attached to one end of a spring makes 20 vibrations in 10 s. That is, 20 vibration in 10 seconds. S in one second it makes 2 vibrations. 
Therefore, the frequency of the object is 2 Hertz.