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# Simple Harmonic Motion

An analysis of Simple Harmonic Motion.

## 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.

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 $\inline&space;-&space;n^2\,x\;\;\;where\;\;\;n^2$ 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 :-

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

This can be expressed as:
$\frac{d}{dx}\left(\frac{1}{2}v^2&space;\right)\;=&space;-&space;n^2\;x$
$\therefore\;\;\;\;\;\frac{1}{2}v^2\;=&space;-&space;\frac{1}{2}n^2\,x^2&space;+&space;Constant$

$\therefore\;\;\;\;\;v^2&space;=&space;n^2(a^2&space;-&space;x^2)$
or,
$v&space;=&space;\pm&space;\;n\sqrt{(a^2&space;-&space;x^2)}$

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

$\therefore\;\;\;\;\;\frac{dx}{dt}&space;=&space;-&space;\;n\sqrt{(a^2&space;-&space;x^2)}$
and,
$\frac{dt}{dx}\;=&space;-&space;\frac{1}{&space;\;n\sqrt{(a^2&space;-&space;x^2)}}$
$\therefore\;\;\;\;\;n\,t&space;=&space;cos^{-1}\frac{x}{a}&space;+&space;K$

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

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

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

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

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

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 $\inline&space;&space;v_0$ when its distance from O is $\inline&space;x_0$, 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 :-

$x&space;=&space;a\cos&space;(nt&space;+&space;\epsilon&space;)$
where $\inline&space;\epsilon$ is a constant]

Now,
$x&space;=&space;x_0\;\;\;when\;\;\;t&space;=&space;0$

$\therefore\;\;\;\;x_0&space;=&space;a&space;\cos&space;\epsilon$
Also $\inline&space;\dot{x}&space;=&space;v_0$ when $\inline&space;t&space;=&space;0$

$\therefore\;\;\;\;v_0=&space;-&space;a\;n&space;\sin&space;\epsilon$
Then,
$a&space;=&space;\sqrt{\left(x_0^2&space;+&space;\frac{v_0^2}{n^2}&space;\right)}$
And,
$a&space;=&space;tan^{-1}\left(-&space;\frac{v_0}{n\;x_0}&space;\right)$

The Constant $\inline&space;\epsilon$ 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 $\inline&space;t&space;+&space;\frac{\epsilon&space;}{n}$ less a multiple of the period.

Also,
$x&space;=&space;a&space;\cos&space;nt&space;\cos&space;\epsilon&space;&space;-&space;a\sin&space;nt&space;\sin&space;\epsilon$
$=&space;x_0\;cos\,nt&space;+&space;\frac{v_o}{n\;sin\,nt}$

In particular if $\inline&space;x_0&space;=&space;0$, i.e. the particle starts from O
$x&space;=&space;\frac{v_0}{n}\;sin\,nt$
and the amplitude is $\inline&space;\frac{v_0}{n}$

Since the acceleration at any instant is
$\ddot{x}$
$\ddot{x}\;=&space;-&space;n^2\,x$

This is characteristic of Simple Harmonic Motion and its solution is given by:-
$x&space;=&space;a\;cos\;(nt&space;+&space;\epsilon&space;)$

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

Let $\inline&space;\epsilon$ 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 $\inline&space;\omega&space;\;t&space;+&space;\epsilon$. Hence, if P is the position of the particle at time t and N the foot of the perpendicular from P on OX, then:-

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

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

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
If the acceleration is $\inline&space;n^2\;x$ at a distance x from the centre then:

$n^2\times2&space;=&space;4\;\;\;\;\;or\;\;\;\;\;n&space;=&space;\sqrt{2}$

Hence the period is:
$\frac{2\;\pi}{n}&space;=&space;\pi\,\sqrt{2}$

If the phase is zero when $\inline&space;t&space;=&space;0$
$x&space;=&space;a\cos&space;n\,t$

where a is the amplitude. Then
$v=&space;-&space;a\,n&space;\sin&space;n\,t$

And the value of v at the centre of oscillation is $\inline&space;\displaystyle&space;\pm&space;a\;n$
$\therefore\;\;\;\;a\;n&space;=&space;8\;ft./sec.$
and
$a&space;=&space;4&space;\sqrt{2}\;ft.$
Solution
The period is
$\frac{2\;\pi}{n}&space;=&space;\pi\,\sqrt{2}$

and amplitude
$a&space;=&space;4&space;\sqrt{2}\;ft.$