**
Simple Harmonic Motion** and **Circular
Motion** are __very__ closely related.
Think about a turntable with a spike attached to it.
Its shadow is projected onto a screen.
As we turn the turntable, we can see the shadow moving forwards and
backwards on the screen

If we look at the apparatus from above:

As **P** goes round
the circumference of the revolving turntable, its projection**
N** will move up the diameter of the circle **AOB**.
The shadow of **P** at this instant is projected onto the screen at **N´**.
The shadow will move up and down the screen along the line **A´N´O´B´**. The turntable is revolving at an angular velocity of w
radians per second, and its linear speed is w*r*
ms^{-1}.

How can we show that the image of **P** describes simple harmonic motion?

Let us consider our shadow going across the screen:

We
know that the acceleration is towards the centre of the circle and is given by *a
= *(2p*f* )^{2}*r*.

Acceleration
is a vector, so has horizontal and vertical components.
We can work out the acceleration parallel to **ONA**
as:

*a = *(2p*f* )^{2
}*r *cos*
*
q

*a = -*(2p*f* )^{2
}*r *cos*
*
q

*x = r *cos* *
q

So
we can combine these two to give:

*a
= - *(2**p f )^{2 }x**

*a
= -Kx*; in this case *K = *(2p*f*
)^{2}.^{2}
» 10.

The
time it takes for our turntable to make **one**
complete revolution is called the **period**,
and is given the code *T*, and is
measured in seconds. It is also the
time for the shadow of **P** to make **one**
**oscillation**, or complete to-and-fro movement.
We can use the simple equation

time (s) = __distance (m)__

speed
(m/s)

and

period (s) = __circumference of the turntable (m)
__

linear speed (m/s)

*T = *__
2____p____r__ =
__2p____r__ =
__2p__

*
v *
w*r *
w

**frequency**, *f
= *1/*T *
Þ
*f* = __
w
__
Þ
**
w
= 2****p f**

2p

**
***
a = - *
w^{2 }*
x*

*
T*
is independent of the radius of the turntable, hence the amplitude of the
oscillation. If the amplitude is
increased, the body travels faster, so *T*
is not affected. This kind of
oscillation is called **isochronous**,
which means that it takes the same time to complete each cycle.

The direction of the **velocity** of anything moving in a circle is always at a **
tangent****:**

So
the component of the velocity parallel to **AOB**
is:

Since
*v* = (2p*f* )*r*,
the velocity parallel to **AOB ***=
-v* sin
q = -(2p*f* )*r*
sin
q

The
negative sign tells us that the velocity is negative when the image is going
upwards and positive when the image is going downwards.
This ties in with the fact that the sine function has positive values for
values of
q between 0 and
p
radians (0 – 180 ^{o}) and negative values from
p to 2p radians (180 ^{o} – 360 ^{o}).

The
derivation of the equation that gives us **velocity**
at any point in the oscillation is rather tedious, but the relationship is:

*v ^{2} =
*(2p

Þ *v ^{2}
= *4p

Þ
*v
= *2**p***f*
Ö**( A^{2} – x^{2})**

In
this relationship, *A* is the **amplitude**
and *s* is the **displacement**
from the equilibrium position. If *x
*= 0, *v* has a maximum value; if *x =
A*, *v* = 0.
The velocity is 0 at each extreme of the oscillation.

We
can easily find the displacement using our circular argument.
If the radius of the turntable is *r*,
we can show quite simply that the displacement, *s*,
if given by:

*
x =
r* cos
q

Since
q =
w*t*,
we can rewrite this as:

*s = r*
cos
w*t*

*f*,
we can rewrite this as:

**
s = **

The plus and minus sign here tells us that the motion is forwards and backwards. Which sign we give for direction is up to the individual. Generally left to right is forwards.