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

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

Displacement on each spring  is same

But restoring force  is different

[latex]F={{F}_{1}}+{{F}_{2}}[/latex]

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

⇒ [latex]-{{k}_{p}}x=-{{k}_{1}}x-{{k}_{2}}x[/latex]

⇒ [latex]-x{{k}_{p}}=-x\left( {{k}_{1}}+{{k}_{2}} \right)[/latex]

⇒ [latex]{{k}_{p}}={{k}_{1}}+{{k}_{2}}[/latex]

Also Read:

Time Period in Parallel Combination

[latex]T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{m}{{{k}_{p}}}}=2\pi \sqrt{\frac{m}{{{k}_{1}}+{{k}_{2}}}}[/latex]

Springs in Series Combination

Springs in Series Combination

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

[latex]x={{x}_{1}}+{{x}_{2}}[/latex]

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

[latex]\frac{F}{-k_{s}}= \frac{F}{-k_{1}}+\frac{F}{-k_{2}}[/latex]

 

[latex]\frac{1}{k_{s}}= \frac{1}{k_{1}}+\frac{1}{k_{2}}[/latex]

⇒ [latex]{{k}_{s}}=\frac{{{k}_{1}}{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}[/latex]

Time Period in Series Combination

[latex]T=2\pi \sqrt{\frac{m}{{{k}_{s}}}}=2\pi \sqrt{\frac{m\left( {{k}_{1}}+{{k}_{2}} \right)}{{{k}_{1}}{{k}_{2}}}}[/latex]

Spring Constant

From Hooke’s law,

Youngs modulus of elasticity, [latex]Y=\frac{Stress}{Strain}=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}[/latex]

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

[latex]\frac{Y\Delta L}{L}=\frac{F}{A}[/latex] [latex]F=\frac{YA}{L}\left( \Delta L \right)[/latex]

Here, [latex]K=\frac{YA}{L}[/latex]

⇒ [latex]K\propto \frac{1}{L}[/latex]

Therefore, the above equation can be written as

⇒ [latex]F=K\,x[/latex]

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

Springs constant k

[latex]K\propto \frac{1}{L}\Rightarrow[/latex] 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

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

[latex]a=\frac{-kx}{m}[/latex]

The acceleration of the particle can be written as

[latex]a=-{{\omega }^{2}}x[/latex] ——(2)

Equating (1) and (2)

[latex]\frac{-k}{m}x=-{{\omega }^{2}}x[/latex]

⇒ [latex]\omega =\sqrt{\frac{k}{m}}[/latex]

Substitute ω value in standard time period expression of SHM

[latex]T=\frac{2\pi }{\omega }=\frac{2\pi }{\sqrt{\frac{k}{m}}}=2\pi \sqrt{\frac{m}{k}}[/latex] [latex]T=2\sqrt{\frac{Mass}{Force\,constant}}[/latex]

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:

[latex]\overrightarrow{a}=-A{{\omega }^{2}}\sin \left( \omega t+\phi \right)[/latex]

⇒ [latex]\overrightarrow{{{a}_{\max }}}=-A{{\omega }^{2}}[/latex]

⇒ [latex]\frac{{{a}_{\max }}}{2}=-\frac{A{{\omega }^{2}}}{2}=-A{{\omega }^{2}}\sin \left( \frac{\pi }{6} \right)[/latex]

Phase [latex]\left( \omega t+\phi \right)=\frac{\pi }{6}[/latex]

⇒ [latex]v=A\omega \cos \left( \frac{\pi }{6} \right)=A\omega \frac{\sqrt{3}}{2}[/latex]

⇒ [latex]x=A\sin \left( \frac{\pi }{6} \right)=\frac{A}{2}[/latex]

⇒ [latex]\left( v=A\omega \frac{\sqrt{3}}{2},\,and\,\,x=\frac{A}{2} \right)[/latex]

 

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:

[latex]v=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}[/latex]

⇒ [latex]{{v}^{2}}={{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)[/latex]

⇒ [latex]\frac{{{v}^{2}}}{{{\omega }^{2}}}=\left( {{A}^{2}}-{{y}^{2}} \right)[/latex]

⇒ [latex]\frac{{{v}^{2}}}{{{\omega }^{2}}}+{{y}^{2}}={{A}^{2}}[/latex] … (1)

⇒ [latex]{{A}^{2}}=\frac{v_{1}^{2}}{{{\omega }^{2}}}+y_{1}^{2}=\frac{v_{2}^{2}}{{{\omega }^{2}}}+y_{2}^{2}[/latex]

⇒ [latex]\frac{v_{1}^{2}-v_{2}^{2}}{{{\omega }^{2}}}=y_{2}^{2}-y_{1}^{2}[/latex]

⇒ [latex]{{\omega }^{2}}=\frac{v_{1}^{2}-v_{2}^{2}}{y_{2}^{2}-y_{1}^{2}}[/latex]

⇒ [latex]\omega =2\pi f[/latex]

⇒ [latex]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}}}[/latex]

 

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:

[latex]KE=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)[/latex]

⇒ [latex]PE=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}[/latex]

⇒ [latex]E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}[/latex]

(a) at [latex]y=\frac{A}{4},[/latex] KE becomes

[latex]KE=\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{\left( \frac{A}{4} \right)}^{2}} \right)[/latex]

= [latex]\frac{15}{16}\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}[/latex]

= [(15/16) x 100] % of  E

= 93% of total energy is KE

(b) KE = PE

[latex]\frac{1}{2}m{{\omega }^{2}}\left( {{A}^{2}}-{{y}^{2}} \right)=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}[/latex] [latex]y=\frac{A}{\sqrt{2}}[/latex]

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) [latex]2\pi \sqrt{\frac{M}{K}}[/latex]

(b) [latex]2\pi \sqrt{\frac{M}{2K}}[/latex]

(c) [latex]2\pi \sqrt{\frac{2M}{3K}}[/latex]

(d) [latex]2\pi \sqrt{\frac{2M}{K}}[/latex]

 

Spring Mass System Solved Examples

Solution:

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

Spring Mass System Solved Examples

[latex]\cos 60{}^\circ =\frac{\Delta x}{x}[/latex]

⇒ [latex]x\cos 60{}^\circ =\Delta \,x[/latex]

⇒ [latex]\frac{x}{2}=\Delta \,x[/latex]

Spring Mass System Solved Examples

[latex]{{F}_{net}}=Kx+2\frac{Kx}{2}\cos 60{}^\circ[/latex]

= [latex]Kx+\frac{Kx}{2}=\frac{3Kx}{2}[/latex]

⇒ [latex]{{K}_{eqn}}x=\frac{3Kx}{2}[/latex]

⇒ [latex]{{K}_{eqn}}=\frac{3K}{2}[/latex]

⇒ [latex]T=2\pi \sqrt{\frac{M}{K}}=2\pi \sqrt{\frac{2M}{3K}}[/latex]

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 [latex]4\times {{10}^{-3}}J[/latex] find the equation of motion of the particle if the initial phase of oscillation is 60°.

(a) [latex]0.1\sin \left( 2t+\frac{\pi }{4} \right)[/latex]

(b) [latex]0.2\sin \left( \frac{1}{2}t+\frac{\pi }{3} \right)[/latex]

(c) [latex]0.2\sin \left( t+\frac{\pi }{3} \right)[/latex]

(d) [latex]0.1\cos \left( 2t+\frac{\pi }{4} \right)[/latex]

Solution:

Equation of motion of particle is

[latex]y=A\sin \left( \omega t+\phi \right)[/latex]

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

From energy

[latex]E=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}[/latex],

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

[latex]{{\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}}[/latex],

[latex]y=0.2\sin \left( t+\frac{\pi }{3} \right)[/latex]

 

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) [latex]\pi s[/latex] (b) [latex]\frac{\pi }{10}s[/latex] (c) [latex]\frac{2\pi }{5}s[/latex] (d) [latex]\frac{\pi }{2}s[/latex]

Solution:

[latex]T=2\pi \sqrt{\frac{M}{K}}=2\pi \sqrt{\frac{0.1}{40}}[/latex]

= [latex]\frac{\pi }{10}s[/latex]

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