Q7.3 The SHM equation (S)

Show that the expression

x=xmcos(ωt+φ)
is a solution to the equation
\[ \ddot{x}=-\omega ^2 x \]

Show that x=xmsin(ωt+φ) is also a solution; explain the relationship between the two solutions.

What do x and ω mean in the context of SHM? What are the units of ω? What can you say about the units of xm?

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Hint

Recall the chain rule for differentiation. Is there a unique answer for the units of xm?
 

Solution

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Solution

Using the chain rule to differentiate

x=xmcos(ωt+φ)
with respect to time gives
\[ \dot{x} =-\omega x_m \sin(\omega t + \phi) \]
and
\[ \ddot{x} =-\omega^2 x_m \cos(\omega t + \phi) = -\omega^2 x \]
Notice that the factors of ω brought out by the operation of differentiation serve to keep the units (dimensions) right.

The argument goes through in exactly the same way if we set x=xmsin(ωt+φ). The fact that we can use either sine or cosine reflects the identity sin(ωt+φ+π/2)=cos(ωt+φ). Thus the ‘cosine’ solution describes exactly the same function as the ‘sine’ solution with the two phase constants differing by π/2.

These are the lynch-pin equations of SHM. In the simplest (and most easily visualised) examples x represents a physical displacement, with units of length. But the ideas and mathematics of SHM are not restricted to such circumstances. One might think of real or model systems in which the oscillating quantity x is a volume, an angle, an electric field, a temperature, a share price or a number of cockroaches. So the units of x are not uniquely defined by this equation: notice that they cancel out. On the other hand ω necessarily always has the units of inverse-time (frequency).