Simple Harmonic Motion | Spring Mass System

Let’s explore different spring mass systems, with mass under linear or rotational motion, and determine angular frequency or time period of simple harmonic motion/oscillations of such a system. More specifically we will look at the following scenarios

Spring Mass System | Linear SHM | Springs in Series

Spring Mass System - Linear SHM - Springs in Series

For springs in series, tension or restoring force exerted by two springs in series is same, i.e. $F = -k_1 x_1 = -k_2 x_2$ or $x_1 + x_2 = x = -F(\cfrac{1}{k_1}+\cfrac{1}{k_2})$
So, equivalent stiffness, $k_{eq} =-\cfrac{F}{x}= \cfrac{k_1 k_2}{k_1+k_2}$
and, angular frequency, $\omega = \sqrt{\cfrac{k_{eq}}{m}}$
and, time period $T = \cfrac{2 \pi}{\omega}$ 

Spring Mass System | Linear SHM | Springs in Parallel

Spring Mass System - Linear SHM - Springs in Parallel

For springs in parallel, the amount of extension, $x$, is the same
and equivalent stiffness $k_{eq} = k_1 + k_2$
so, angular frequency, $\omega = \sqrt{\cfrac{k_{eq}}{m}}$ 

Spring Mass System | Linear SHM | Multiple Springs in 2D Arrangement

Spring Mass System - Linear SHM - Multiple Springs in 2D Arrangement

Given that the mass is allowed to move along the $y-$axis (dotted line),
For a small displacement, $\delta$, of mass $m$ (from its mean position) along say positive $y$-axis, the restoring force, $F$ will be $=(k_1 + 2 k_2 \cos^2 \theta) \delta$
So, $k_{eq} = k_1 + 2 k_2 \cos^2 \theta$
and angular frequency, $\omega = \sqrt{\cfrac{k_{eq}}{m}}$ 

Spring Mass System | 2D SHM

Spring Mass System - 2D SHM​

If $k_1 = k_2$, then the mass will move along a straight line

If $k_1 = 4k_2$, meaning $\omega_1 = 2 \omega_2$ then the particle will trace a horizontal 8 as shown. Note that as the particle completes 1 oscillation in the horizontal direction, it completes two oscillations in the vertical direction
(and if $4k_1 = k_2$, then the particle will trace the standing or vertical 8)

You can use similar approach to determine the resulting trajectory of the particle for other combinations of $k_1$ and $k_2$.

Spring Mass System | Linear SHM | Pulley - Spring - Mass System

Spring Mass System - Linear SHM - Pulley - Spring - Mass System

Assuming that the pulleys are massless, let’s say that we displace the mass $m$ by a small distance $\delta$ in the downward direction (from its mean or equilibrium position), let’s figure out the increase (or change) in tension.
so, let’s say that the increase in tension is $T$,
then
Increase in stretching of spring $1$, $\delta_1 = \cfrac{2T}{k_1}$
Similarly, increase in stretching of spring $2$ will be $\delta_2 = \cfrac{2T}{k_2}$
$\delta_3 = \cfrac{2T}{k_3}$
$\delta_4 = \cfrac{2T}{k_4}$

And $2 \delta_1 + 2 \delta_2 + 2 \delta_3 + 2 \delta_4 = \delta$ 

So, $4T (\cfrac{1}{k_1} + \cfrac{1}{k_2} + \cfrac{1}{k_3} + \cfrac{1}{k_4}) = \delta$

Now, $k_{eq} = \cfrac{T}{\delta}$
i.e. $k_{eq} = \cfrac{1}{4(\cfrac{1}{k_1} + \cfrac{1}{k_2} + \cfrac{1}{k_3} + \cfrac{1}{k_4})}$

& angular frequency $\omega = \sqrt{\cfrac{k_{eq}}{m}}$

Spring Mass System | Linear SHM | Constant Force

Spring Mass System - Linear SHM - Constant Force

Note that, in the equilibrium position spring is compressed by $\cfrac{mg \sin \theta}{k}$ where $\theta$ is the angle of incline, assuming the incline surface is frictionless

Now if we displace mass $m$ slightly from its mean or equilibrium position, it will oscillate with angular frequency $\omega = \sqrt{\cfrac{k}{m}}$ 

Linear SHM | Two particles executing SHM

Linear SHM - Two particles executing SHM

If we displace masses $m_1$ and $m_2$ (connected by spring of spring constant $k_{eq}$) towards each other and then release them, they will undergo simple harmonic motion (with their center of mass at rest)

So, let’s write the newton’s law $F = ma = m \cfrac{d^2 x}{dt^2}$ for mass $m_1$, when it is a distance $\delta_1$ from its mean position towards left

Well to determine the restoring spring force $F$, we need to know the total extension of the spring when $m_1$ is at $\delta_1$ to the left of its mean position.

Well recall that $x_{cm} = \cfrac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

Now if you take origin at center of mass then $m_1 x_1 + m_2 x_2 = 0$ or $m_1 \delta_1 + m_2 \delta_2 = 0$ or $\delta_2 = – \cfrac{m_1}{m_2} \delta_1$ with minus sign indicating that if $\delta_1$ was towards left $\delta_2$ will be towards right

so net extension of the spring $=\delta_1 + \delta_2 = \cfrac{m_1 + m_2}{m_2} \delta_1$

or, $-k_{eq} \cfrac{m_1 + m_2}{m_2} \delta_1$ $=m_1 \cfrac{d^2 \delta_1}{dt^2}$

or $\cfrac{d^2 \delta_1}{dt^2} = – \cfrac{k_{eq}}{\cfrac{m_1 m_2}{m_1+m_2}} \delta_1$

i.e. $\omega^2 = \cfrac{k_{eq}}{\cfrac{m_1 m_2}{m_1+m_2}}$ 

Angular SHM | Scenario 1

Angular SHM - Scenario 1

If we rotate the rod by a small angle $\theta$ about its center, the springs will be stretched by $\cfrac{L}{2} \theta$

So, the restoring torque about the pivot point will be $\tau = \cfrac{L}{2}$$(k_1 \cfrac{L}{2} \theta +$ $k_2 \cfrac{L}{2} \theta)$ $=\cfrac{L^2}{4} (k_1 + k_2) \theta$

Now, as per newton’s 2nd law $\tau = I \alpha = I \cfrac{d^2 \theta}{dt^2}$

So, $\cfrac{d^2 \theta}{dt^2}=-\cfrac{L^2}{4I} (k_1 + k_2) \theta$, where $I = \cfrac{1}{12} ML^2$ for the rod

So angular frequency, $\omega^2 = \cfrac{L^2}{4I} (k_1 + k_2)$

Note that above is true only for small angular displacements. For larger angular displacements the motion will not be simple harmonic 

Angular SHM | Scenario 2

Angular SHM - Scenario 2

Note that assuming that the springs are ideal springs, motion will be simple harmonic ir-respective of how large the angular displacement, $\theta$, is from its mean equilibrium position of $PQ$

Here the restoring force on each mass is $2 k_1 R \theta + 2 k_2 R \theta$ $=2(k_1 + k_2) R \theta$

So, restoring torque on mass $m$ is $2(k_1 + k_2) R^2 \theta$

Now, writing netwon’s 2nd law for mass $m$, $\tau = I \alpha$ $= I \cfrac{d^2 \theta}{dt^2}$ $mR^2 \cfrac{d^2 \theta}{dt^2}$

i.e. $\cfrac{d^2 \theta}{dt^2} = – \cfrac{2(k_1+k_2)}{m} \theta$ 

or, angular frequency $\omega = \sqrt{\cfrac{2(k_1+k_2)}{m}}$

Simple Harmonic Motion | Linear + Rotational Motion

Simple Harmonic Motion - Linear + Rotational Motion

Assuming that cylinder is in pure rolling at all times, let’s say that the center of cylinder is displaced by a distance $x$ to the right as shown, then the net restoring force acting on it is

$F_{net} = -(3kx – f_s)$ (assuming that the friction force is acting to the right)

Newton’s 2nd law ($F_{net}=Ma_{cm}$) gives us
$-(3kx – f_s) = M a_{cm}$ $\cdots$(i)

Now torque on cylinder about its center of mass $\tau_{net} =-(2kx+f_s)R$

Using newton’s 2nd law $\tau_{net} = I \alpha$ we get
$-(2kx+f_s)R = I \alpha$

Now in pure rolling $a_{cm} = \alpha R$
i.e. $-(2kx+f_s) = a_{cm} \cfrac{I}{R^2}$ $\cdots$(ii)

Solving for $a_{cm}$ in terms of $x$ we get
$a_{cm} = \cfrac{d^2x}{dt^2}=$ $-\cfrac{5k}{M + \cfrac{I}{R^2}} x$

or angular frequency $\omega = \sqrt{\cfrac{5k}{M + \cfrac{I}{R^2}}}$ 

This Post Has One Comment

  1. Souvik Sinha

    solution for the last question is wrong. If COM moves by distance x, then top point will move by distance 2x because the top point has double velocity of the COM i.e Vcom +rw=v+v=2v. so after that , if we calculate then w=sqrt(5k/M+I/R^2), which is correct because if we solve the problem by Instantaneous center approach we also get the same result. Here ground contact point is the IC (instantaneous centre).

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