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.
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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](https://latex.codecogs.com/svg.image?\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 :-
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
![\inline cos^{-1}](https://latex.codecogs.com/svg.image?\inline&space;&space;cos^{-1})
= 0 and hence K = 0
and,
When
![\inline nt=\n\frac{\pi}{2}](https://latex.codecogs.com/svg.image?\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 \frac{\pi}{2n}](https://latex.codecogs.com/svg.image?\inline&space;\frac{\pi}{2n})
with a velocity of
![\inline -an](https://latex.codecogs.com/svg.image?\inline&space;-an)
. It will continue along the straight line and its velocity will be zero when
![\inline nt = \pi](https://latex.codecogs.com/svg.image?\inline&space;nt&space;=&space;\pi)
and
![\inline x = - a](https://latex.codecogs.com/svg.image?\inline&space;x&space;=&space;-&space;a)
. It will then return to O arriving when
![\inline nt = \frac{3}{2}\pi](https://latex.codecogs.com/svg.image?\inline&space;nt&space;=&space;\frac{3}{2}\pi)
with a velocity
![\inline an](https://latex.codecogs.com/svg.image?\inline&space;an)
and reach A in a time
![\inline \frac{2\pi}{n}](https://latex.codecogs.com/svg.image?\inline&space;&space;\frac{2\pi}{n})
with zero velocity. The motion is then repeated indefinitely unless destroyed by some force.
Note:
- The time
is called the Period of the oscillation and is the time for one complete cycle. - If the frequency is f and the period
, then
.
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](https://latex.codecogs.com/svg.image?\inline&space;&space;v_0)
when its distance from O is
![\inline x_0](https://latex.codecogs.com/svg.image?\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 :-
where
![\inline \epsilon](https://latex.codecogs.com/svg.image?\inline&space;\epsilon)
is a constant]
Now,
Also
![\inline \dot{x} = v_0](https://latex.codecogs.com/svg.image?\inline&space;\dot{x}&space;=&space;v_0)
when
Then,
And,
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The Constant
![\inline \epsilon](https://latex.codecogs.com/svg.image?\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 t + \frac{\epsilon }{n}](https://latex.codecogs.com/svg.image?\inline&space;t&space;+&space;\frac{\epsilon&space;}{n})
less a multiple of the period.
Also,
In particular if
![\inline x_0 = 0](https://latex.codecogs.com/svg.image?\inline&space;x_0&space;=&space;0)
, i.e. the particle starts from O
and the amplitude is
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
![\inline a](https://latex.codecogs.com/svg.image?\inline&space;a)
with uniform angular velocity
![\inline \omega](https://latex.codecogs.com/svg.image?\inline&space;\omega)
, its orthogonal projection on a diameter of the circle moves on the diameter in simple harmonic motion of amplitude
![\inline a](https://latex.codecogs.com/svg.image?\inline&space;a)
and period
![\inline \frac{2\;\pi}{\omega }](https://latex.codecogs.com/svg.image?\inline&space;\frac{2\;\pi}{\omega&space;})
.
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Let
![\inline \epsilon](https://latex.codecogs.com/svg.image?\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 \omega \;t + \epsilon](https://latex.codecogs.com/svg.image?\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:-
and the point moves with Simple Harmonic Motion of amplitude
![\inline a](https://latex.codecogs.com/svg.image?\inline&space;a)
and period
[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
![](https://latex.codecogs.com/svg.image?\inline&space;n^2\;x)
at a distance
x from the centre then:
Hence the period is:
If the phase is zero when
where
a is the amplitude. Then
And the value of
v at the centre of oscillation is
and
Solution
The period is
and amplitude