 # Spring Mass System

## What is Spring Mass System?

A spring-mass system in simple terms can be described as a spring sytem where a block is hung or attached at the free end of the spring. The spring-mass system can usually be used to find the period of any object performing the simple harmonic motion. The spring-mass system can also be used in a wide variety of applications. For instance, spring-mass system can be used to simulate the motion of human tendons using computer graphics as well as foot skin deformation. ### How Does Mass Affect the Period of a Spring?

Consider a spring with mass m with spring constant k, in a closed environment spring demonstrates a simple harmonic motion.

T = 2π √m/k

From the above equation, it is clear that the period of oscillation is free from both gravitational acceleration and amplitude. Also, a constant force cannot alter the period of oscillation. Meanwhile, the time period is directly proportional to the mass of the body that is fixed to the spring. When a heavy object is attached to it, it will oscillate more slowly.

## Spring Mass System Arrangements

Spring mass systems can be arranged in two ways. These include;

• The parallel combination of springs

• Series combination of springs

We will discuss them below;

### Parallel Combination of Springs Fig (a), (b) and (c) – are the parallel combination of springs.

Displacement on each spring is the same.

But restoring force  is different;

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

Since, F = -kx, the above equation can be written as

$-{{k}_{p}}x=-{{k}_{1}}x-{{k}_{2}}x$

$-x{{k}_{p}}=-x\left( {{k}_{1}}+{{k}_{2}} \right)$

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

### Time Period in Parallel Combination

$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{{{k}_{p}}}}=2\pi \sqrt{\frac{m}{{{k}_{1}}+{{k}_{2}}}}$

### Springs in Series Combination The force on each string is the same but displacement on each string is different

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

Since, F = -kx, the above equation can be written as

$\frac{F}{-k_{s}}= \frac{F}{-k_{1}}+\frac{F}{-k_{2}}$

$\frac{1}{k_{s}}= \frac{1}{k_{1}}+\frac{1}{k_{2}}$

${{k}_{s}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}$

### Time Period in Series Combination

$T=2\pi \sqrt{\frac{m}{{{k}_{s}}}}=2\pi \sqrt{\frac{m\left( {{k}_{1}}+{{k}_{2}} \right)}{{{k}_{1}}{{k}_{2}}}}$

## Spring Constant

From Hooke’s law,

Youngs modulus of elasticity, $Y=\frac{Stress}{Strain}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$

Here,

F = Force needed to extend or compress the spring

A = area over which the force is applied

L = nominal length of the material

ΔL = change in the length

$\frac{Y\Delta L}{L}=\frac{F}{A}$ $F=\frac{YA}{L}\left( \Delta L \right)$

Here, $K=\frac{YA}{L}$

$K\propto \frac{1}{L}$

Therefore, the above equation can be written as

$F=K\,x$

If a spring of spring constant (K) and length (L) cutted into $\frac{L}{2}$ size two pieces, then magnitude of spring constant of the new pieces will be? $K\propto \frac{1}{L}\Rightarrow$ then K becomes = 2K for the new pieces.

## Spring Constant Video

### Understanding Spring Constant ## How to Find the Time period of a Spring Mass System? Steps:

1. Find the mean position of the SHM (point at which Fnet = 0) in horizontal spring-mass system

The natural length of the spring = is the position of the equilibrium point.

2. Displace the object by a small distance (x) from its equilibrium position (or) mean position . The restoring force for the displacement ‘x’ is given as

F=-k{x}
—–(1)

The acceleration of the body is given as a = F/m

Substituting the value of F from equa (1) we get

$a=\frac{-kx}{m}$

The acceleration of the particle can be written as

$a=-{{\omega }^{2}}x$ ——(2)

Equating (1) and (2)

$\frac{-k}{m}x=-{{\omega }^{2}}x$

⇒ $\omega =\sqrt{\frac{k}{m}}$

Substitute ω value in standard time period expression of SHM

$T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{k}{m}}}=2\pi \sqrt{\frac{m}{k}}$ $T=2\sqrt{\frac{Mass}{Force\,constant}}$

## Problems on Spring Mass System

Q.1: A particle is executing linear SHM what are its velocity and displacement when its acceleration is half the maximum possible?

Solution:

$\overrightarrow{a}=-A{{\omega }^{2}}\sin \left( \omega t+\phi \right)$

$\overrightarrow{{{a}_{\max }}}=-A{{\omega }^{2}}$

$\frac{{{a}_{\max }}}{2}=-\frac{A{{\omega }^{2}}}{2}=-A{{\omega }^{2}}\sin \left( \frac{\pi }{6} \right)$

Phase $\left( \omega t+\phi \right)=\frac{\pi }{6}$

$v=A\omega \cos \left( \frac{\pi }{6} \right)=A\omega \frac{\sqrt{3}}{2}$

$x=A\sin \left( \frac{\pi }{6} \right)=\frac{A}{2}$

$\left( v=A\omega \frac{\sqrt{3}}{2},\,and\,\,x=\frac{A}{2} \right)$

Q.2: A particle executing linear SHM has speeds v1 and v2 at distances y1 and y2 from the equilibrium position. What is the frequency of the oscillation of the particle?

Solution:

$v=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}$

${{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)$

$\frac{{{v}^{2}}}{{{\omega }^{2}}}=\left( {{A}^{2}}-{{y}^{2}} \right)$

$\frac{{{v}^{2}}}{{{\omega }^{2}}}+{{y}^{2}}={{A}^{2}}$ … (1)

${{A}^{2}}=\frac{v_{1}^{2}}{{{\omega }^{2}}}+y_{1}^{2}=\frac{v_{2}^{2}}{{{\omega }^{2}}}+y_{2}^{2}$

$\frac{v_{1}^{2}-v_{2}^{2}}{{{\omega }^{2}}}=y_{2}^{2}-y_{1}^{2}$

${{\omega }^{2}}=\frac{v_{1}^{2}-v_{2}^{2}}{y_{2}^{2}-y_{1}^{2}}$

$\omega =2\pi f$

$f=\frac{\omega }{2\pi }=\frac{1}{2\pi }{{\left[ \frac{v_{1}^{2}-v_{2}^{2}}{y_{2}^{2}-y_{1}^{2}} \right]}^{\frac{1}{2}}}$

Q.3: A particle is executing SHM of amplitude A.

(a) What fraction of the total energy is kinetic when displacement is a quarter of the amplitude?

(b) At what displacement is the energy are half kinetic and half potential?

Solution:

$KE=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)$

$PE=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}$

$E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}$

(a) at $y=\frac{A}{4},$ KE becomes

$KE=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{\left( \frac{A}{4} \right)}^{2}} \right)$

= $\frac{15}{16}\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}$

= [(15/16) x 100] % of  E

= 93% of total energy is KE

(b) KE = PE

$\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}$ $y=\frac{A}{\sqrt{2}}$

Q.4: Three springs each of force constant k are connected at equal angles with respect to each other to a common mass. If the mass is pulled by anyone of the spring then the time period of its oscillation?

(a) $2\pi \sqrt{\frac{M}{K}}$

(b) $2\pi \sqrt{\frac{M}{2K}}$

(c) $2\pi \sqrt{\frac{2M}{3K}}$

(d) $2\pi \sqrt{\frac{2M}{K}}$ Solution:

It is pulled by an upper spring each are making equal angles. $\cos 60{}^\circ =\frac{\Delta x}{x}$

$x\cos 60{}^\circ =\Delta \,x$

$\frac{x}{2}=\Delta \,x$ ${{F}_{net}}=Kx+2\frac{Kx}{2}\cos 60{}^\circ$

= $Kx+\frac{Kx}{2}=\frac{3Kx}{2}$

${{K}_{eqn}}x=\frac{3Kx}{2}$

${{K}_{eqn}}=\frac{3K}{2}$

$T=2\pi \sqrt{\frac{M}{K}}=2\pi \sqrt{\frac{2M}{3K}}$

Q.5: A particle of mass 0.2 kg is executing SHM of amplitude 0.2 m. When the particle passes through the mean position. It mechanical energy is $4\times {{10}^{-3}}J$ find the equation of motion of the particle if the initial phase of oscillation is 60°.

(a) $0.1\sin \left( 2t+\frac{\pi }{4} \right)$

(b) $0.2\sin \left( \frac{1}{2}t+\frac{\pi }{3} \right)$

(c) $0.2\sin \left( t+\frac{\pi }{3} \right)$

(d) $0.1\cos \left( 2t+\frac{\pi }{4} \right)$

Solution:

Equation of motion of particle is

$y=A\sin \left( \omega t+\phi \right)$

A = 0.2 m, $\omega =?,\,\,\phi =60{}^\circ ,\,\,ME=4\times {{10}^{-3}}J$

From energy

$E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}$,

$4\times {{10}^{-3}}=\frac{1}{2}\left( 0.2 \right){{\omega }^{2}}{{\left( 0.2 \right)}^{2}}$,

${{\omega }^{2}}=\frac{4\times {{10}^{-3}}\times 2}{\left( 0.2 \right){{\left( 0.2 \right)}^{2}}}=\frac{8\times {{10}^{-3}}}{0.008}=1\,rad\,{{s}^{-1}}$,

$y=0.2\sin \left( t+\frac{\pi }{3} \right)$

Q.6: A block of mass 0.1 kg which slides without friction on a 30° incline is connected to the top of the incline by a massless spring of force constant 40 Nm-1. If the block is pulled slightly from its mean position what is the period of oscillation? (a) $\pi s$ (b) $\frac{\pi }{10}s$ (c) $\frac{2\pi }{5}s$ (d) $\frac{\pi }{2}s$

Solution:

$T=2\pi \sqrt{\frac{M}{K}}=2\pi \sqrt{\frac{0.1}{40}}$

= $\frac{\pi }{10}s$

Test your Knowledge on Spring mass system