Simple Harmonic Motion

An analysis of Simple Harmonic Motion.

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Contents

Simple Harmonic Motion.
Other Initial Conditions
The Relation To Uniform Motion In A Circle.
Page Comments

Simple Harmonic Motion.

If a particle moves in a straight line in such a way that its acceleration is always directed towards a fixed point on the line and is proportional to the distance from the point, the particle is said to be moving in Simple Harmonic Motion.

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Let O be a fixed point on the line X'X and x the distance of the particle from O at any time t. Also, let the acceleration of the particle along OX as $\inline - n^2\,x\;\;\;where\;\;\;n^2$ is a positive constant. No matter whether x is positive or negative the acceleration will be directed towards O.

If v is the velocity at time t, the acceleration will be in the direction OX and will be given by the differential relationship :-

$v\;\frac{dv}{dx} = -n^2\,x$

(2)

This can be expressed as:

$\frac{d}{dx}\left(\frac{1}{2}v^2 \right)\;= - n^2\;x$

(3)

$\therefore\;\;\;\;\;\frac{1}{2}v^2\;= - \frac{1}{2}n^2\,x^2 + Constant$

(4)

$\therefore\;\;\;\;\;v^2 = n^2(a^2 - x^2)$

(5)

or,

$v = \pm \;n\sqrt{(a^2 - x^2)}$

(6)

In the initial stage of the motion v is negative as the particle is moving towards O

$\therefore\;\;\;\;\;\frac{dx}{dt} = - \;n\sqrt{(a^2 - x^2)}$

(7)

and,

$\frac{dt}{dx}\;= - \frac{1}{ \;n\sqrt{(a^2 - x^2)}}$

(8)

$\therefore\;\;\;\;\;n\,t = cos^{-1}\frac{x}{a} + K$

(9)

When t = 0 x = a and $\inline cos^{-1}$ = 0 and hence K = 0

$\therefore\;\;\;\;\;n\,t = cos^{-1}\frac{x}{a}$

(10)

and,

$x = a \cos nt$

(11)

$\therefore\;\;\;\;\;v= - a\,n\sin nt$

(12)

When $\inline nt=\n\frac{\pi}{2}$ cos nt = 0, and thus a particle starting from A moving towards O arrives in a time $\inline \frac{\pi}{2n}$ with a velocity of $\inline -an$ . It will continue along the straight line and its velocity will be zero when $\inline nt = \pi$ and $\inline x = - a$ . It will then return to O arriving when $\inline nt = \frac{3}{2}\pi$ with a velocity $\inline an$ and reach A in a time $\inline \frac{2\pi}{n}$ with zero velocity. The motion is then repeated indefinitely unless destroyed by some force.

Note:

The time $\inline \frac{2\;\pi}{n}$ is called the Period of the oscillation and is the time for one complete cycle.
If the frequency is f and the period $\inline \frac {2\;\pi}{n}$ , then $\inline f = \frac{n}{2\pi}$ .

Also,

If the period of the motion is known, the motion is completely determined.
The Period maybe written down at once if the magnitude of the acceleration for some value of x is known.
The amplitude is determined by the initial displacement.

Other Initial Conditions

If the the motion is started by giving the particle a velocity $\inline v_0$ when its distance from O is $\inline x_0$ , the type of motion is unchanged and the time is measured from this instant, instead of the instant when x = a. In this case the value of x at any instant is given by :-

$x = a\cos (nt + \epsilon )$

(13)

where $\inline \epsilon$ is a constant]

Now,

$x = x_0\;\;\;when\;\;\;t = 0$

(14)

$\therefore\;\;\;\;x_0 = a \cos \epsilon$

(15)

Also $\inline \dot{x} = v_0$ when $\inline t = 0$

$\therefore\;\;\;\;v_0= - a\;n \sin \epsilon$

(16)

Then,

$a = \sqrt{\left(x_0^2 + \frac{v_0^2}{n^2} \right)}$

(17)

And,

$a = tan^{-1}\left(- \frac{v_0}{n\;x_0} \right)$

(18)

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The Constant $\inline \epsilon$ is called the Epoch of the motion. The Phase of the motion at time t is the time which has elapsed since the particle was at the positive end of its path. Thus the phase is $\inline t + \frac{\epsilon }{n}$ less a multiple of the period.

Also,

$x = a \cos nt \cos \epsilon - a\sin nt \sin \epsilon$

(19)

$= x_0\;cos\,nt + \frac{v_o}{n\;sin\,nt}$

(20)

In particular if $\inline x_0 = 0$ , i.e. the particle starts from O

$x = \frac{v_0}{n}\;sin\,nt$

(21)

and the amplitude is $\inline \frac{v_0}{n}$

Since the acceleration at any instant is

$\ddot{x}$

(22)

$\ddot{x}\;= - n^2\,x$

(23)

This is characteristic of Simple Harmonic Motion and its solution is given by:-

$x = a\;cos\;(nt + \epsilon )$

(24)

may be written down if the amplitude and the epoch are known.

The Relation To Uniform Motion In A Circle.

If a particle is describing a circle of radius $\inline a$ with uniform angular velocity $\inline \omega$ , its orthogonal projection on a diameter of the circle moves on the diameter in simple harmonic motion of amplitude $\inline a$ and period $\inline \frac{2\;\pi}{\omega }$ .

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Let $\inline \epsilon$ be the angle which the radius initially makes with the diameter X'OX . Then after a time t the angle made by the radius to the particle is $\inline \omega \;t + \epsilon$ . Hence, if P is the position of the particle at time t and N the foot of the perpendicular from P on OX, then:-

$ON = a \cos (\omega \,t + \epsilon )$

(25)

and the point moves with Simple Harmonic Motion of amplitude $\inline a$ and period $\inline \frac{2\;\pi}{\omega }$

Example:

[imperial]

Example - Simple Harmonic Motion

Problem

A particle moves with Simple Harmonic Motion in a straight line. Find the time of a complete oscillation if the acceleration is 4 ft/sec², when the distance from the centre of the oscillation is 2 ft. If the Velocity with which the particle passes through the centre of oscillations is 8 ft./sec. find the amplitude.

Workings