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 $- n^2\,x\;\;\;where\;\;\;n^2$ 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)

or,

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

(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 $nt=\frac{\pi}{2}$ 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 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 $v_0$ 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,

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

(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)

$$= x_0\;cos\,nt + \frac{v_0}{n}\;sin\,nt$$

(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 [imperial]

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/sec\^2, 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 $n^2\;x$ at a distance x from the centre then:

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

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

If the phase is zero when $t = 0$

$$x = a\cos n\,t$$

(25)

where a is the amplitude. Then

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

(26)

And the value of v at the centre of oscillation is $\displaystyle \pm a\;n$ $\therefore\;\;\;\;a\;n = 8\;ft./sec.$ and $a = 4 \sqrt{2}\;ft.$

Solution

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

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

Worked example #857 not found.