Spring Mass System

What is Spring Mass System?

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.

Parallel Combination of Springs

Parallel Combination of Springs Parallel Combination of Springs

Parallel Combination of Springs

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

Displacement on each spring = is same

But restoring force = is different

F=F1+F2F={{F}_{1}}+{{F}_{2}}

kpx=k1xk2x-{{k}_{p}}x=-{{k}_{1}}x-{{k}_{2}}x

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

kp=k1+k2{{k}_{p}}={{k}_{1}}+{{k}_{2}}

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

Force on each string is same but displacement on each string is different

x=x1+x2x={{x}_{1}}+{{x}_{2}} F1=k1x1{{F}_{1}}=-{{k}_{1}}{{x}_{1}}

T=2πLgeff=2πLg+aT=2\pi \sqrt{\frac{L}{{{g}_{eff}}}}=2\pi \sqrt{\frac{L}{g+a}} F2=k2x2{{F}_{2}}=-{{k}_{2}}{{x}_{2}}

1ks=1k1+1k2\frac{1}{{{k}_{s}}}=\frac{1}{{{k}_{1}}}+\frac{1}{{{k}_{2}}} Fk1=x1\frac{F}{{{k}_{1}}}={{x}_{1}}

ks=k1k2k1+k2{{k}_{s}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}} Fk2=x2\frac{F}{{{k}_{2}}}={{x}_{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

K=YALK=\frac{YA}{L} Y = youngs modulus of elasticity

From Hooke’s law

Y=StressStrain=FADLLY=\frac{Stress}{Strain}=\frac{\frac{F}{A}}{\frac{DL}{L}} YDLL=FA\frac{YDL}{L}=\frac{F}{A} F=YAL(DL)F=\frac{YA}{L}\left( DL \right) [Since,  K=YAL]\left[ Since, \;K=\frac{YA}{L} \right]

F=KxF=K\,x

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

K1LK\propto \frac{1}{L}

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 Spring length (L)

K1LK\propto \frac{1}{L}\Rightarrow then K becomes = 2K for the new pieces.

How to Find the Time period of a Spring Mass System?

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 equilibrium point.

2. Displace the object by a small distance (x) from its equilibrium position (or) mean position other than mean position, restoring force will act on the body

Fnet=kx\overrightarrow{{{F}_{net}}}=-k\overrightarrow{x}

a=kmx\overrightarrow{a}=\frac{-k}{m}\overrightarrow{x}

3. Acceleration of the particle is calculated and the calculated value of a\overrightarrow{a}.

ax\overrightarrow{a}\propto -\overrightarrow{x} then only it is SHM

Then equate the a=ω2x\overrightarrow{a}=-{{\omega }^{2}}\overrightarrow{x} kmx=ω2x\frac{-k}{m}\overrightarrow{x}=-{{\omega }^{2}}\overrightarrow{x}

ω=km\omega =\sqrt{\frac{k}{m}}

4. 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=2InertiaForceconstantT=2\sqrt{\frac{Inertia}{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:

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?

Solution:

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 quarter of the amplitude?

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

(a) 9090%,\,\,\frac{A}{\sqrt{2}}

(b) 9494%,\,\,\frac{A}{\sqrt{3}}

(c) 99%,\,\,\frac{A}{\sqrt{4}}

(d) 9393%,\,\,\frac{A}{\sqrt{2}}

Solution:

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}}

Spring Mass System Solved Examples

(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}}

= 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}}

Solution:

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

Spring Mass System Solved Examples 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

Fnet

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}

Keqnx=3Kx2{{K}_{eqn}}x=\frac{3Kx}{2}

Keqn=3K2{{K}_{eqn}}=\frac{3K}{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)

Solution:

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

Solution:

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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