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An analysis of Simple Harmonic Motion

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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 with 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 $- n^2\,x\;\;\;where\;\;\;n^2$ .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 :-

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

(1)

This can be expressed as:

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

(2)

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

(3)

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

(4)

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

(5)

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

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

(6)

and

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

(7)

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

(8)

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

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

(9)

and

$x = a \cos nt$

(10)

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

(11)

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

Note.

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

Also

If the period of the motion are 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 $v_0$ when its distance from O is $x_0$ , 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 :-

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

(12)

where $\epsilon$ is a constant]

Now

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

(13)

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

(14)

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

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

(15)

Then

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

(16)

And

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

(17)

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

Also

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

(18)

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

(19)

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

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

(20)

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

Since the acceleration at any instant is

$\ddot{x}$

(21)

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

(22)

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

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

(23)

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

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

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

(24)

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

Example 1

A particle moves with Simple Harmonic Motion in a straight line. Find the time of a complete oscillation if the acceleration is 4 ft./sec sqd. 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.

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

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

(25)

Hence the period is:

$\frac{2\;\pi}{n} = \pi\,\sqrt{2}$

(26)

If the phase is zero when $t = 0$

$x = a\cos n\,t$

(27)

where "a" is the amplitude. Then

$v= - a\,n \sin n\,t$

(28)

And the value of v at the centre of oscillation is $\displaystyle \pm a\;n$

$\therefore\;\;\;\;a\;n = 8\;ft./sec.$

(29)

and

$a = 4 \sqrt{2}\;ft.$

(30)

Forces Causing Simple Harmonic Motion

When a body moves in a straight line under the action of a force which is directed towards a fixed point O on the line and which is proportional to the distance x of the body from the point.|The body will move with Simple Harmonic Motion. If the Force is $\mu\,x$ the the Force in the direction of an increase in x is $- \mu\,x$ . This is negative when x is positive and visa versa.

If W is the weight of the body it follows that by Newton's Second Law, the equation of motion is :-

$\mathbf{\frac{W}{g}\;\ddot{x}= - \mu \,x}$

(31)

i.e. $\ddot{x} + n^2\,x = 0$ where $n^2 = \frac{\mu \,g}{W}$

This equation is of the form:

$\frac{d^2x}{dt^2} + n^2y = 0$

(32)

The Auxiliary equation is

$m^2 + n^2 = 0$

(33)

and hence $m = \pm \,j\,n$

$\therefore\;\;\;\;\;\;x = A\,cos\,nt + B\,sin\,nt$

(34)

This equation can also be written as:

$\displaystyle x = r \sin(nt + \alpha )$

Whatever the initial conditions, x is a periodic function of t, of period $\displaystyle \frac{2\;\pi }{n}$ . As $\displaystyle n^2 = \frac{\mu \;g}{W}$ the period depends only upon the weight of the body and $\displaystyle \mu$ and does not depend upon the amplitude r of the oscillations.

Example 2

If the weight of a body is 16 lbs. and $\displaystyle \mu$ = 18lbs./ft. find the period of its oscillation. Also, if the body is given a velocity of 1/3 ft./sec. away from O find the displacement t secs. later and the amplitude of the oscillation.

Applying Newton's Second Law

$\frac{W}{g}\ddot{x}\;= - \mu \;x$

(35)

i.e

$\frac{16}{32}\;\ddot{x}= - 18x$

(36)

$\therefore\;\;\;\;\ddot{x} + 36\;x = 0$

(37)

$\therefore\;\;\;\;\;x = A\cos 6t + B\sin 6t$

(38)

Hence the Period is $\displaystyle \frac{2\,\pi }{6}\approx 1.05\;secs.$

$\dot{x}= - 6\,A\sin 6t + 6\,B\cos 6t$

(39)

But at t = 0 $\displaystyle x = \frac{1}{2\times12}$ and $\dot{x} = \frac{1}{3}$

Substituting in equation (8)

$\frac{1}{24} = A$

(40)

And in equation (9)

$\frac{1}{3} = 6\,B$

(41)

$\therefore\;\;\;\;x = \frac{1}{24}\cos 6t + \frac{1}{18}\sin 6t$

(42)

$=\frac{5}{72}\sin(6t + \alpha )$

(43)

where $\tan \alpha = \frac{3}{4}$

Thus the amplitude of the oscillations is

$\frac{5}{72}\;ft.$

(44)

Example 3

In a brake test on an engine the coefficient of friction between the rope and the wheel is $\mu$ . The rope carries a load W at one end and is attached to a spring of stiffness s at the other. Find the period of oscillation of W, the general expression for the dis[placement of W in terms of t and the least and greatest values for the tension of the rope on the side of W when W is making oscillations of amplitude a.

When the spring is extended a distance x ,its tension on the other side of the wheel is $e^{\mu\,\pi}\times s\,x$

Applying Newton's Second Law

$\displaystyle \frac{W}{g}\ddot{x} = W - e^{\mu\,\pi}\times s\,x$

From which it can be shown that:

$\ddot{x} + \frac{e^{\mu\pi}\,s\,g}{W}\left(x - \frac{W\,e^{-\mu\pi}}{s} \right)} = 0$

(45)

Hence the weight oscillates with a period which is given by :

$2\pi \sqrt{\frac{W}{e^{\mu\pi}s\,g}}$

(46)

about a centre at a distance $\displaystyle \frac{W\,e^{-\mu\pi}}{s}$

The general expression for x in terms of t is :

$x = \frac{W\,e^{-\mu\pi}}{s} + a\;sin\,(nt + \alpha)$

(47)

where

$n = \sqrt{\frac{e^{\mu\pi}\,s\,g}{W}}$

(48)

The greatest and least tensions of the rope are the values of $e^{\mu\pi}\,s\,g$ when W is in its lowest and highest positions. i.e. when $\displaystyle x = \frac{W\,e^{-\mu\pi}}{s}\;\pm a$ . The greatest and least value tensions are therefore $\displaystyle W\;\pm e^{\mu\pi}\;s\,a$

If the period of oscillation of W is 1/3 secs. and a = 1/4 in then $e^{\mu\pi}\;s\;\approx\;\frac{9\pi^3}{8}W$ and $e^{\mu\pi\;s\,a\;\approx0.23W$ .

Therefore the least and greatest tensions are 0.77 W and 1.23W

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Please Note. A second collection of worked examples will be published very shortly. A number are based on Simple Harmonic Motion

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