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.

Vertical spring-mass system

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

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;


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


xkp=x(k1+k2)-x{{k}_{p}}=-x\left( {{k}_{1}}+{{k}_{2}} \right)


Also Read:

Time Period in Parallel Combination

T=2πω=2πmkp=2πmk1+k2T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{{{k}_{p}}}}=2\pi \sqrt{\frac{m}{{{k}_{1}}+{{k}_{2}}}}

Springs in Series Combination

Springs in Series Combination

The force on each string is the same but displacement on each string is different


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

Fks=Fk1+Fk2\frac{F}{-k_{s}}= \frac{F}{-k_{1}}+\frac{F}{-k_{2}}


1ks=1k1+1k2\frac{1}{k_{s}}= \frac{1}{k_{1}}+\frac{1}{k_{2}}


Time Period in Series Combination

T=2πmks=2πm(k1+k2)k1k2T=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=StressStrain=FAΔLLY=\frac{Stress}{Strain}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}


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

YΔLL=FA\frac{Y\Delta L}{L}=\frac{F}{A} F=YAL(ΔL)F=\frac{YA}{L}\left( \Delta L \right)

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

K1LK\propto \frac{1}{L}

Therefore, the above equation can be written as


If a spring of spring constant (K) and length (L) cutted into L2\frac{L}{2} size two pieces, then magnitude of spring constant of the new pieces will be?

Springs constant k

K1LK\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?

Spring Mass System


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


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

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


The acceleration of the particle can be written as

a=ω2xa=-{{\omega }^{2}}x ——(2)

Equating (1) and (2)

kmx=ω2x\frac{-k}{m}x=-{{\omega }^{2}}x

⇒ ω=km\omega =\sqrt{\frac{k}{m}}

Substitute ω value in standard time period expression of SHM

T=2πω=2πkm=2πmkT=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{k}{m}}}=2\pi \sqrt{\frac{m}{k}} T=2MassForceconstantT=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?


a=Aω2sin(ωt+ϕ)\overrightarrow{a}=-A{{\omega }^{2}}\sin \left( \omega t+\phi \right)

amax=Aω2\overrightarrow{{{a}_{\max }}}=-A{{\omega }^{2}}

amax2=Aω22=Aω2sin(π6)\frac{{{a}_{\max }}}{2}=-\frac{A{{\omega }^{2}}}{2}=-A{{\omega }^{2}}\sin \left( \frac{\pi }{6} \right)

Phase (ωt+ϕ)=π6\left( \omega t+\phi \right)=\frac{\pi }{6}

v=Aωcos(π6)=Aω32v=A\omega \cos \left( \frac{\pi }{6} \right)=A\omega \frac{\sqrt{3}}{2}

x=Asin(π6)=A2x=A\sin \left( \frac{\pi }{6} \right)=\frac{A}{2}

(v=Aω32,andx=A2)\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?


v=ωA2y2v=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}

v2=ω2(A2y2){{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)

v2ω2=(A2y2)\frac{{{v}^{2}}}{{{\omega }^{2}}}=\left( {{A}^{2}}-{{y}^{2}} \right)

v2ω2+y2=A2\frac{{{v}^{2}}}{{{\omega }^{2}}}+{{y}^{2}}={{A}^{2}} … (1)

A2=v12ω2+y12=v22ω2+y22{{A}^{2}}=\frac{v_{1}^{2}}{{{\omega }^{2}}}+y_{1}^{2}=\frac{v_{2}^{2}}{{{\omega }^{2}}}+y_{2}^{2}

v12v22ω2=y22y12\frac{v_{1}^{2}-v_{2}^{2}}{{{\omega }^{2}}}=y_{2}^{2}-y_{1}^{2}

ω2=v12v22y22y12{{\omega }^{2}}=\frac{v_{1}^{2}-v_{2}^{2}}{y_{2}^{2}-y_{1}^{2}}

ω=2πf\omega =2\pi f

f=ω2π=12π[v12v22y22y12]12f=\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?


KE=12mω2(A2y2)KE=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)

PE=12mω2y2PE=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}

E=12mω2A2E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}

(a) at y=A4,y=\frac{A}{4}, KE becomes

KE=12mω2(A2(A4)2)KE=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{\left( \frac{A}{4} \right)}^{2}} \right)

= 151612mω2A2\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

12mω2(A2y2)=12mω2y2\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}} y=A2y=\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) 2πMK2\pi \sqrt{\frac{M}{K}}

(b) 2πM2K2\pi \sqrt{\frac{M}{2K}}

(c) 2π2M3K2\pi \sqrt{\frac{2M}{3K}}

(d) 2π2MK2\pi \sqrt{\frac{2M}{K}}


Spring Mass System Solved Examples


It is pulled by an upper spring each are making equal angles.

Spring Mass System Solved Examples

cos60=Δxx\cos 60{}^\circ =\frac{\Delta x}{x}

xcos60=Δxx\cos 60{}^\circ =\Delta \,x

x2=Δx\frac{x}{2}=\Delta \,x

Spring Mass System Solved Examples

Fnet=Kx+2Kx2cos60{{F}_{net}}=Kx+2\frac{Kx}{2}\cos 60{}^\circ

= Kx+Kx2=3Kx2Kx+\frac{Kx}{2}=\frac{3Kx}{2}



T=2πMK=2π2M3KT=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 4×103J4\times {{10}^{-3}}J find the equation of motion of the particle if the initial phase of oscillation is 60°.

(a) 0.1sin(2t+π4)0.1\sin \left( 2t+\frac{\pi }{4} \right)

(b) 0.2sin(12t+π3)0.2\sin \left( \frac{1}{2}t+\frac{\pi }{3} \right)

(c) 0.2sin(t+π3)0.2\sin \left( t+\frac{\pi }{3} \right)

(d) 0.1cos(2t+π4)0.1\cos \left( 2t+\frac{\pi }{4} \right)


Equation of motion of particle is

y=Asin(ωt+ϕ)y=A\sin \left( \omega t+\phi \right)

A = 0.2 m, ω=?,ϕ=60,ME=4×103J\omega =?,\,\,\phi =60{}^\circ ,\,\,ME=4\times {{10}^{-3}}J

From energy

E=12mω2A2E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}},

4×103=12(0.2)ω2(0.2)24\times {{10}^{-3}}=\frac{1}{2}\left( 0.2 \right){{\omega }^{2}}{{\left( 0.2 \right)}^{2}},

ω2=4×103×2(0.2)(0.2)2=8×1030.008=1rads1{{\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=0.2sin(t+π3)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?

Spring Mass System Solved Examples

(a) πs\pi s (b) π10s\frac{\pi }{10}s (c) 2π5s\frac{2\pi }{5}s (d) π2s\frac{\pi }{2}s


T=2πMK=2π0.140T=2\pi \sqrt{\frac{M}{K}}=2\pi \sqrt{\frac{0.1}{40}}

= π10s\frac{\pi }{10}s

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