# A-level Physics (Advancing Physics)/Simple Harmonic Motion

Simple harmonic motion occurs when the force on an object is proportional and in the opposite direction to the displacement of the object. Examples include masses on springs and pendula, which 'bounce' back and forth repeatedly. Mathematically, this can be written:
$F=-kx$,
where F is force, x is displacement, and k is a positive constant. This is exactly the same as Hooke's Law, which states that the force F on an object at the end of a spring equals -kx, where k is the spring constant. Since F = ma, and acceleration is the second derivative of displacement with respect to time t:
$m{\frac {d^{2}x}{dt^{2}}}=-kx$
${\frac {d^{2}x}{dt^{2}}}={\frac {-kx}{m}}$
The solution of this second order differential equation is:
$x=A\cos {\omega t}$,
where A is the maximum displacement, and ω is the 'angular velocity' of the object. The derivation is given here, since it will seem very scary to those who haven't met complex numbers before. It should be noted that this solution, if given different starting conditions, becomes:
$x=A\sin {\omega t}$,

## Angular Velocity

Angular velocity in circular motion is the rate of change of angle. It is measured in radians per. second. Since 2π radians is equivalent to one complete rotation in time period T:
$\omega ={\frac {2\pi }{T}}=2\pi f$
If we substitute this into the equation for displacement in simple harmonic motion:
$x=A\cos {2\pi ft}$
The reason the equation includes angular velocity is that simple harmonic motion is very similar to circular motion. If you look at an object going round in a circle side-on, it looks exactly like simple harmonic motion. We have already noted that a mass on a spring undergoes simple harmonic motion. The following diagram shows the similarity between circular motion and simple harmonic motion:

## Time Period

The time period of an oscillation is the time taken to repeat the pattern of motion once. In general:
$T={\frac {2\pi }{\omega }}$
However, depending on the type of oscillation, the value of ω changes. For a mass on a spring:
$\omega ={\sqrt {\frac {k}{m}}}$
For a pendulum:
$\omega ={\sqrt {\frac {g}{l}}}$,
where g is the gravitational field strength, and l is the length of the string. By substitution, we may gain the following table:
| Type of Oscillation | Spring | Pendulum |
| Angular Velocity | ${\sqrt {\frac {k}{m}}}$ | ${\sqrt {\frac {g}{l}}}$ |
| Time Period | $2\pi {\sqrt {\frac {m}{k}}}$ | $2\pi {\sqrt {\frac {l}{g}}}$ |

## Velocity and Acceleration

The displacement of a simple harmonic oscillator is:
$x=A\cos {\omega t}$
Velocity is the rate of change of displacement, so:
$v={\frac {dx}{dt}}=-A\omega \sin {\omega t}$
Acceleration is the rate of change of velocity, so:
$a={\frac {dv}{dt}}={\frac {d^{2}x}{dt^{2}}}=-A\omega ^{2}cos{\omega t}=-\omega ^{2}x$

## Questions

1. A 10N weight extends a spring by 5 cm. Another 10N weight is added, and the spring extends another 5 cm. What is the spring constant of the spring?
2. The spring is taken into outer space, and is stretched 10 cm with the two weights attached. What is the time period of its oscillation?
3. What force is acting on the spring after 1 second? In what direction?
4. A pendulum oscillates with a frequency of 0.5 Hz. What is the length of the pendulum?
5. The following graph shows the displacement of a simple harmonic oscillator. Draw graphs of its velocity, momentum, acceleration and the force acting on it.
6. A pendulum can only be modelled as a simple harmonic oscillator if the angle over which it oscillates is small. Why is this?
Worked Solutions