# Derivation of the Simple Harmonic Motion Equation

This page shows how the equation (or rather proportionality) for the equation for the S.H.M. of a spring/trolley oscillator system can be derived. Its level is not meant to be above that expected at higher level GCSE/first year of A-Level.

*The preliminary
data are as follows:*

*Consider a trolley
connected between two sets of springs of different spring constants.
In both arrangements the trolley has the same mass and the same amplitude
of oscillation is used. The period in (a) is twice that of (b) and therefore
(b) oscillates at twice the speed of (a).*

From this information we can see the following:

*x*, we can therefore say that in one complete oscillation both trolleys move

*4x*.

Trolley (a) T

Trolley (b) T/2

(taking the period of (a) as being T, the period of (b) must be half).Trolley (a)

Trolley (b)

From this we can say that Trolley (b) has twice the speed of (a).Trolley (a) T/4

Trolley (b)

*v*, we can say that the maximum speed of trolley (b) is twice that of (a), as it covers the same distance in half the time. This is therefore

*2v*. From this we can calculate the average acceleration of the trolleys:

Trolley (a)

Trolley (b)

Which can be simplified to give:

Trolley (a)

Trolley (b)

Therefore (b) has 4 times the average acceleration of (a).

*F=ma*, therefore as

*m*is constant

*F*is directly proportional to

*a*. This then means that if a is increased four times, F is also increased by the same factor. We can therefore say that:

(Where these are the forces acting on the respective trolleys).

*x*, (the amplitude of oscillation) is constant,

*F*is directly proportional to

*k*, therefore if

*F*is increased by a factor of 4, (as in the example above),

*k*must have increased by the same factor. From this we can see that the spring in (b) must have four times the Hooke's constant of the spring in (a).

*k*quadruples the period halves. This therefore suggests that: