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Simple Harmonic Motion

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 in Simple Harmonic Motion.

23287/Simple-Harmonic-Motion-100.png
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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 - 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 :-

This can be expressed as:

or,

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

and,

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

and,

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 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 :-

where \epsilon is a constant]

Now,

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

Then, And,

23287/Simple-Harmonic-Motion-103.png
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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,

In particular if x_0 = 0, i.e. the particle starts from O and the amplitude is \frac{v_0}{n}

Since the acceleration at any instant is

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

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 }.

23287/Simple-Harmonic-Motion-101.png
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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:-

and the point moves with Simple Harmonic Motion of amplitude a and period \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
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

where a is the amplitude. Then

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.