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\left( {{k}_{1}}+{{k}_{2}} \right)}{{{k}_{1}}{{k}_{2}}}}\end{array})

Spring Constant

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

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

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( \Delta L \right)\end{array})

(\begin{array}{l}K=\frac{Y}{A} \div 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^2x}{dt^2}…(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 $\left( \omega t+\phi \right)=\frac{\pi }{6}$

(\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 \left( v=\frac{A\omega \sqrt{3}}{2},,and,,x=\frac{A}{2} \right) \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}}\left( {{y}^{2}}-{{A}^{2}} \right)\end{array})

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

‘(\begin{array}{l}PE = \frac{1}{2}m\omega^2y^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}}\left( {{A}^{2}}-{{\left( \frac{A}{4} \right)}^{2}} \right)\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{align*}y &= \frac{A}{\sqrt{2}}\end{align*})’

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}\Delta ,x\end{array} )

(\begin{array}{l}\frac{x}{2}=\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: $E = 4 × 10^{-3} J$ and $x(t) = 0.2 \sin(2\pi ft + 60^{\circ})$

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

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

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

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

Given:

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Solution:

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Equation of motion for the particle is

(\begin{array}{l}y = A \sin\left(\omega t + \phi\right)\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

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

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

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

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

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})

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