What is a Spring Mass System?

A spring-mass system is a type of system that can be used to calculate the period of an object undergoing simple harmonic motion. Additionally, it can be used for a wide range of applications, such as simulating the motion of human tendons with computer graphics and foot skin deformation.

What is the Relationship Between Mass and the Period of a Spring?

Consider a spring with mass m and spring constant k in a closed environment, which demonstrates Simple Harmonic Motion (SHM).

$$T=\frac{2\pi \sqrt{m}}{k}$$

From the above equation, it is evident that the period of oscillation is independent of both gravitational acceleration and amplitude. Additionally, a constant force cannot affect the period of oscillation. Furthermore, the time period is directly proportional to the mass of the body that is attached to the spring. Therefore, when a heavy object is connected to it, it will oscillate more slowly.

Spring Mass System Arrangements

• Spring mass systems can be arranged in two ways: 1. 2.

The parallel combination of springs

Series Combination of Springs

We will discuss them below;

Parallel Combination of Springs

Fig (a), (b), and (c) are parallel combinations of springs.

Displacement on each spring is equal.

But restoring force is different;

$$\begin{array}{l}F=F_1+F_2\end{array}$$

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

$$\begin{array}{l}-k_{p}x=-k_{1}x-k_{2}x\end{array}$$ \Rightarrow

\Rightarrow -x{{k}_{p}} = -x{{k}_{1}} - x{{k}_{2}}

$$\begin{array}{l}\Rightarrow k_p=k_1\cdot k_2\end{array}$$

Check Out:

Important Concepts of Simple Pendulum

Newton’s Second Law of Motion

Free Body Diagram

Parallel Combination of Time Periods

($$\begin{array}{l}T=2\pi \sqrt{\frac{m}{k_1+k_2}}=2\pi \sqrt{\frac{m}{k_p}}=\frac{2\pi }{\omega }\end{array}$$ )

Springs in Series Combination

The force on each string is the same, but the displacement of each string is different.

($$\begin{array}{l}{x}_1+x_2=x\end{array}$$)

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

($$\begin{array}{l}\frac{F}{k_s}=\frac{F}{k_1}+\frac{F}{k_2}\end{array}$$ )

($$\begin{array}{l}{k}_s=k_1{k}_2\\ \frac{1}{k_s}=\frac{1}{k_1}+\frac{1}{k_2}\end{array}$$)

$$\Rightarrow k_s=\frac{k_1\cdot k_2}{k_1+k_2}$$

Time Period in Series Combination

($$\begin{array}{l}T=2\pi \sqrt{\frac{m(k_1+k_2)}{k_1{k}_2}}\end{array}$$)

Spring Constant

The relationship between force and displacement described by Hooke’s Law

Young’s Modulus of Elasticity, $$Y=\frac{Stress}{Strain}=\frac{\frac{F}{A}}{\frac{\mathrm{\Delta }L}{L}}$$

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})

($$\begin{array}{l}F=\frac{Y}{L}\left(\mathrm{\Delta }L\right)\end{array}$$)

($$\begin{array}{l}K=\frac{Y}{A}÷L\end{array}$$)

($$\begin{array}{l}K\propto \frac{1}{L}\end{array}$$)

Therefore, the equation can be rewritten as:

‘($$\begin{array}{l}F=K\cdot x\end{array}$$)’

The magnitude of spring constant of the new pieces will be 2K.

($$\begin{array}{l}K\propto \frac{1}{L}\end{array}$$ )

so, (K = \frac{2K}{L})

Spring Constant: A Video

Understanding the 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 (where the net force is equal to 0) in a horizontal spring-mass system.

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

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 = -kx (1)

The acceleration of the body is given as a = $\frac{F}{m}$

Substituting the value of F from equation (1), we get

($$\begin{array}{l}a=\frac{kx}{m}\end{array}$$)

The acceleration of the particle can be expressed as

($$\begin{array}{l}a=\frac{d^{2}x}{d{t}^2}\dots (2)\end{array}$$ )

Equating (1) and (2)

(\frac{k}{m} = {{\omega }^{2}})

‘$\omega =\sqrt{\frac{k}{m}}$’

Substitute the value of ω in the standard time period expression of Simple Harmonic Motion.

($$\begin{array}{l}T=2\pi \sqrt{\frac{m}{k}}\end{array}$$)

$$T = 2\sqrt{\frac{Mass}{Force\,constant}}$$

Problems on Spring-Mass Systems

Q.1: The velocity and displacement of a particle executing linear SHM when its acceleration is half the maximum possible is zero.

Given:

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Solution:

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(\overrightarrow{a} = A \omega^2 \cos(\omega t + \phi))

‘$\overrightarrow{a_{max}}=A{\omega }^2$’

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

Phase $(\omega t+\varphi )=\frac{\pi }{6}$

($$\begin{array}{l}\Rightarrow v=A\omega \cos \left(\frac{\pi }{6}\right)=A\omega \frac{\sqrt{3}}{2}\end{array}$$)

($$\begin{array}{l}\Rightarrow x=A\sin \left(\frac{\pi }{6}\right)=\frac{A}{2}\end{array}$$)

($$\begin{array}{l}\Rightarrow (v=\frac{A\omega \sqrt{3}}{2},,and,,x=\frac{A}{2})\end{array}$$)

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

Given:

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Solution:

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\(\begin{array}{l}v=\omega \sqrt{{{A}^{2}}-{{y}^{2}}} \end{array}\)

($$\begin{array}{l}\Rightarrow v^2={\omega }^2(y^2-A^2)\end{array}$$)

($$\begin{array}{l}\Rightarrow \frac{{\omega }^2}{v^2}=(y^2-A^2)\end{array}$$)

($$\begin{array}{l}\frac{v^2}{{\omega }^2}+y^2=A^2\\ Rightarrow(1)\end{array}$$)

$$A^2=\frac{v_1^2}{{\omega }^2}+y_1^2=\frac{v_2^2}{{\omega }^2}+y_2^2$$

($$\begin{array}{l}\frac{v_2^2-v_1^2}{{\omega }^2}=y_1^2-y_2^2\end{array}$$ )

($$\begin{array}{l}\Rightarrow \frac{v_1^2-v_2^2}{y_2^2-y_1^2}={\omega }^2\end{array}$$ )

‘$\omega =2\pi f$’

$$f=\frac{\omega }{2\pi }=\frac{1}{2\pi }{\left[\frac{v_1^2-v_2^2}{y_2^2-y_1^2}\right]}^{\frac{1}{2}}$$

Q.3: What is the maximum displacement of the particle?

What fraction of the total energy is kinetic when the displacement is one-fourth of the amplitude?

At what displacement is the energy half kinetic and half potential?

Given:

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Solution:

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

‘($$\begin{array}{l}PE=\frac{1}{2}m{\omega }^2{y}^2\end{array}$$)’

($$\begin{array}{l}\Rightarrow E=\frac{1}{2}m{\omega }^2{A}^2\end{array}$$)

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

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

\(\frac{15}{32}m\omega^2A^2\)

100% of E = [(15/16) x 100]%

93% of total energy is Kinetic Energy

KE = PE

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

‘($$\begin{array}{rl}y& =\frac{A}{\sqrt{2}}\end{array}$$)’

Q.4: What is the time period of oscillation of the common mass when it is pulled by one of the three springs, each of which has a force constant k and are connected at equal angles with respect to each other?

\(\begin{array}{l}(a)\ 2\pi \sqrt{\frac{K}{M}}\end{array} \)

\(\displaystyle 2\pi \sqrt{\frac{M}{2K}}\)

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

\(\displaystyle 2\pi \sqrt{\frac{2M}{K}}\)

Given:

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Solution:

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

(\cos 60{}^\circ = \frac{x}{\Delta x})

($$\begin{array}{l}\Rightarrow x\cos 60{}^{\circ }=\frac{\sqrt{3}}{2}\mathrm{\Delta },x\end{array}$$ )

($$\begin{array}{l}\frac{x}{2}=\mathrm{\Delta },x\end{array}$$)

‘$\begin{array}{l}{F}_{net}=Kx+Kx\cos 60{}^{\circ }\end{array}$’

($$\begin{array}{l}Kx+\frac{Kx}{2}=\frac{3Kx}{2}\end{array}$$ \Rightarrow 2Kx = 3Kx \Rightarrow Kx = 0\end{array})

($$\begin{array}{l}{K}_{eqn}x=\frac{2Kx}{3}\end{array}$$)

($$\begin{array}{l}\Rightarrow K_{eqn}=\frac{3K}{2}\end{array}$$)

‘($$\begin{array}{l}T=2\pi \sqrt{\frac{2M}{3K}}\end{array}$$ )’

Q.5: The equation of motion of the particle with mass 0.2 kg, executing SHM of amplitude 0.2 m and initial phase of oscillation of 60°, when passing through the mean position, is given by: $E=4\times {10}^{-3}J$ and $x(t)=0.2\sin (2\pi ft+{60}^{\circ })$

($$\begin{array}{l}(a)0.1\sin (2t+\frac{\pi }{4})\end{array}$$)

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

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

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

Given:

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Solution:

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Equation of motion for the particle is

($$\begin{array}{l}y=A\sin (\omega t+\varphi )\end{array}$$)

A = 0.2 m, (\omega = \frac{ME}{A^2} = \frac{4\times {{10}^{-3}}J}{(0.2 m)^2} ), (\phi = 60{}^\circ ), ME = 4 x 10\textsuperscript{-3} J

To energy

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

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

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

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

Q.6: What is the period of oscillation of a 0.1 kg block sliding without friction on a 30° incline, which is connected to the top of the incline by a massless spring of force constant 40 Nm-1, when it is pulled slightly from its mean position?

(a) $\pi$ s

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

2π/5 s

(d) $\frac{\pi }{2}$ s

Given:

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Solution:

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($$\begin{array}{l}T=2\pi \sqrt{\frac{0.1}{40}}=0.09817477\end{array}$$)

\(\frac{\pi}{10}s\)