The Physics of Simple Harmonic Motion

An object moving in an SHM must obey the relation

\begin{aligned} a=-\omega ^2 x\end{aligned}

where x is the displacement and \omega is the angular frequency of the oscillator. \omega  is related to the frequency f of the oscillator by the relation \begin{aligned} w=2\pi f \end{aligned} .

A body is moving in SHM if its acceleration is directly proportional to the displacement and in the opposite direction to the displacement.

In this equation, we observe the following:

  • the acceleration of the oscillator is always opposite to the displacement,
  • the magnitude of the acceleration is directly proportional to the displacement.

From these two points, we see that the acceleration is greatest when the oscillator is at the maximum displacement while it is zero when the oscillator is at equilibrium. Since force and acceleration is directly proportional F=ma , we see that the restoring force of the oscillator is correspondingly largest at the maximum displacement and zero at equilibrium.

SHM in a spring

When a object hung on a spring oscillates, it is moving in SHM. There is a restoring force from the spring causing the object to move up and down. There are two ways to start the motion:

  1. displace the object to a certain distance and release, or
  2. with the oscillator at equilibrium, give it a push such that it starts moving.

Which colour of the graphs would case 1 and 2 belong?


The red graph would have an equation of \begin{aligned} x=x_0 \sin\omega t \end{aligned} while the blue graph is \begin{aligned} x=x_0 \cos \omega t \end{aligned}.

For the rest of this post, I will use the first equation (red), \begin{aligned} x = x_0 \sin \omega t \end{aligned}.

Since velocity is the derivative of displacement with time,

\begin{aligned} v=\frac{dx}{dt}=x_0 \omega \cos \omega t  \end{aligned}

Acceleration is the derivative of velocity with time, hence

\begin{aligned} a=\frac{dv}{dt}= x_0 \omega^2 \sin \omega t\end{aligned}

But from the first equation x = x_0 \sin \omega t , we have

\begin{aligned} a = -x_0 \omega^2 \end{aligned}

Equation relating velocity to displacement

In the above equations, we have relationships with respect to time. We can also determine the velocity of the object if the displacement is known.

From the original displacement-time equation,

\begin{aligned} x&=x_0 \sin{\omega t} \end{aligned}

\begin{aligned} v=x_0 \omega \cos{\omega t} \end{aligned}

Squaring both equations, we have

\begin{aligned} x^2&=x_0 ^2 \sin^2{\omega t}\\v^2&=x_0 ^2 \omega ^2\cos^2{\omega t}\\&=x_0 ^2 \omega^2 (1-\sin^2{\omega t})\\&=x_0 ^2 \omega^2(1-\frac{x^2}{x_0 ^2})\end{aligned}


\begin{aligned} v^2 &= \omega^2(x_0 ^2 - x^2) \end{aligned}


\begin{aligned} v = \pm\omega \sqrt(x_0 ^2 - x^2) \end{aligned}

The Restoring Force

It is important to understand that the restoring force is the resultant of both the spring force and the weight of the oscillator. Since weight is constant, the restoring force would be proportional to the spring force, F=-kx. Hence, the restoring force is never constant.

\begin{aligned} F_R=F_S-w \end{aligned}

where F_R is the restoring force and F_S is the spring force.

From the equation, it can be seen that the maximum restoring force occurs when F_S is maximum, which occurs when the displacement is at maximum. This is also the position where the acceleration is maximum(and zero velocity).

Phases of an SHM

\begin{aligned} x &= x_0 \sin \omega t \\ v &= x_0 \omega \cos \omega t\\ a &= -x_0 \omega ^2 \sin \omega t\end{aligned}

As we can see,

  • the velocity of SHM v and displacement x has a phase difference of \begin{aligned} \frac{\pi}{2} \end{aligned} .
  • The acceleration a and the displacement has a phase difference of \pi.
  • The acceleration and velocity has a phase difference of \frac{\pi}{2}

Review Question 1

Using the blue graph above, sketch the corresponding graphs of v and a.