##
Further
Mechanics Tutorial 8 - Energy
in SHM

We
have seen how we can work out the velocity at any given point by *v*^{2}
= (2p*f*
)^{2}(*A*^{2}* – x*^{2}).
We also know that** kinetic energy**, *E*_{k}
= ˝ *mv*^{2}, so it doesn’t
take a genius to see that:

*
E*_{k}
=
˝
*m*(2p*f* )^{2}(*A*^{2}* – x*^{2})

We
can see that the maximum **potential energy**
is found at each end of the swing. As
the pendulum swings away from the rest position, work is done against the
restoring force. If *x*
= 0, the equilibrium position, the restoring force is 0 as well.
At any displacement *s* we can
find the restoring force *F*.
We can do this by using Newton II, *F
= ma* and *a = *(2p*f* )^{2}*x*.

So
the force at any displacement *s* is
given by *F = m*(2p*f*
)^{2}*x*.

We
know that work done = force
´
distance moved. However the force
is not constant and we need to consider the **average
force****, **which is half the
maximum force.**
**

Work done = average force
´
displacement = ˝ *m*(2p*f*
)^{2}*x*
´
*x* = ˝ *m*(2p*f*
)^{2}*x*^{2}.

So
we can work out the potential energy at any point using:

*
E*_{p}
=
˝ *m*(2**p***f***
)**^{2}*x*^{2}*
*

The
**total energy** at any point is simply
the sum of the potential and kinetic energy;

*
E*_{tot} = E_{p} + E_{k}
*
*

*=
*˝ *m*(2p*f* )^{2}(*A*^{2}* – x*^{2})
+ ˝ *m*(2p*f* )^{2}*x*^{2}

= **˝ ***m*(2**p***f* )^{2}*A*^{2}