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Simple Harmonic Motion

An analysis of the velocity and acceleration of a particle in simple harmonic motion.

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Contents

Overview
Velocity And Acceleration In Simple Harmonic Motion
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Overview

Key facts

For a particle undergoing simple harmonic motion, with the displacement $s$ is given by:

$s = m \cos nt$

with $m,n=ct$ .
Its squared velocity is:

$v^2 = n^2(m^2-s^2)$

And its acceleration:

$a = -n^2s$

Simple harmonic motion is the periodic motion of a particle which follows a sinusoidal oscillation about an equilibrium position, demonstrating a single resonant frequency. It is illustrated by the motion of an object on a spring when subjected to an elastic restoring force. When the object is displaced from its equilibrium position, it experiences a net restoring force which causes it to accelerate back to the original position. As it moves closer to equilibrium the restoring force decreases, becoming zero exactly at the equilibrium position. The momentum of the object carries it past the equilibrium position, and the restoring force once again tends to slow it down, until its velocity dissappears.

In simple harmonic motion the acceleration is always proportional to the displacement from the equilibrium position, and acts in a direction opposite to it.

Velocity And Acceleration In Simple Harmonic Motion

In order to define the velocity and acceleration of a simple harmonic motion, we introduce the displacement $s$ with the following equation:

$s = m \cos nt$

(1)

where $m$ and $n$ are constants.

Taking into account the definition of velocity (for a more detailed discussion on velocity and acceleration see Linear Velocity and Acceleration ) and equation (1), we can write:

$v = \frac{ds}{dt} = -mn \sin nt$

(2)

We can express $v$ from (2) only in terms of $s$ by squaring the equation, which gives:

$v^2 = m^2n^2\sin^2 nt = m^2n^2(1-\cos^2 nt) = n^2(m^2-m^2\cos^2 nt)$

from which, by also considering (1), we obtain the squared velocity:

$v^2 = n^2(m^2-s^2)$

(3)

By differentiating (3) with respect to $s$ , we get:

$2v \frac{dv}{ds} = n^2(-2s)$

(4)

Considering one of the forms in which the acceleration can be expressed, and using (4) we obtain:

$a = v\frac{dv}{ds} = \frac{n^2(-2s)}{2}$

from which:

$a = -n^2s$

(5)

The same equation for the acceleration of a simple harmonic motion can also be obtained by considering an alternate expression form for acceleration, and using equation (2), we have:

$a = \frac{dv}{dt} = -mn^2 \cos nt$

(6)

Coupling (6) with (1) leads again to the final equation for the acceleration highlighted in (5).