Interactive textbook: Oscillations


It would be impossible to write an equation for the motion of a fly buzzing about a room (or a hummingbird about a woodland). There needs to be some pattern in the movement that we can replicate mathematically. Motion with constant acceleration has a simple pattern that can be modelled using linear and quadratic equations. 

Oscillations are more complex but if we simplify the situation we can see a pattern that can be modelled using a sinusoidal function. Physicists are good at spotting patterns and developing mathematical models to make predictions.

This is why a lot of the best physics students are head-hunted by banks as they leave university!

Understanding the model

To use a model to make predictions it is important that you don't just accept the model. Instead, you must seek to understand it... but you can't understand a model if you don't know where it has come from. Although the SL course does not require you to know the derivations of the oscillations equations, but they have been included anyway. The mathematics isn't that complicated so you should be able to follow the steps (even though you needn't memorise them). For a complete understanding of oscillations you need to use calculus. You may not have studied calculus in maths yet, so in this chapter we have used alternative methods. If you do calculus you should return to this topic and the following will all make sense.

The topic in a nutshell?

If \(x=x_0\sin(\omega t)\)...

Then \({\mathrm{d}^2x\over\mathrm{d}t^2}=-\omega^2x_0\sin(\omega t)\).


The area under an acceleration - time graph gives the:

If a is constant area = aΔt = v - u


Simple pendulum

The simple pendulum is a small mass hanging on a light inextensible string. Its motion can be represented on a displacement vs time graph. If the mass is undisturbed it will hang vertically; we will take this position to represent zero displacement. The motion is 2-dimensional so it will have displacement in the \(x\) and \(y\) directions. Here we will consider only the \(x\)-displacement.

This looks like a sine curve but it's not quite...

By reducing the size of the swing we get a closer fit.

There is something special about the sine function. If you plot the gradient of the sine function you get a cosine function and if you plot the gradient of the cosinefunction you get an inverted sine function. So if the graph of displacement vs time is a sine function, the graph of velocity vs time will be a cosine function and the graph of acceleration vs time will be a -sine function.

If you do HL Mathematics you will learn about calculus and come to appreciate this using differentiation:

If \(y=\sin(x)\)...

Then \({\mathrm{d}y\over\mathrm{d}x}=\cos(x)\)...

And \({\mathrm{d}^2y\over\mathrm{d}x^2}=-\sin(x)\).

The sine and cosine curves are the same shape but displaced along the \(x\)-axis. Whether displacement is a sine function or a cosine function depends on the starting point. In the example below we will start from maximum positive displacement so the displacement vs time curve is a cosine curve. This is quite likely to match reality, as a pendulum is usually released from the top of the swing.


The graph represents an oscillation.

Which graph represents the velocity vs time?

velocity is the gradient of s - t


Which of these graphs would represent the acceleration vs time?

Acceleration is zero apart from at the points where it changes, here it is infinite. +1 to -1 is negative, -1 to +1 is positive.

Note that the acceleration is not proportional to displacement, this is only true for sinusoidaly varying dispacement.



Half an oscillation in stages:

  • The ball starts on the right with positive displacement and is set to swing through the middle to the left. The gradient at the start is zero so the velocity is zero. 
  • As the ball swings to the left it has negative increasing velocity, shown by the line on the graph with increasing negative gradient.
  • As the ball passes the centre it has maximum negative velocity so the gradient is maximum and negative. As the ball passes this position of zero displacement its displacement becomes negative.
  • The ball then starts to slow down so the gradient decreases, reaching zero at maximum negative displacement.

x equals x subscript o cos left parenthesis 2 pi f t right parenthesis equals x subscript o cos omega t


Half an oscillation in stages:

  • The ball starts at rest so velocity is zero.
  • It then has an increasing velocity to the left, which means it has negative acceleration. The acceleration is the gradient of this graph so we can see that the acceleration decreases as the ball swings to the middle.
  • At the centre the velocity has its maximum value but the gradient is zero so acceleration is zero.
  • As the ball passes the centre the gradient turns positive so the acceleration is now positive and increasing to a maximum at the top of the swing.

v equals negative v subscript o sin omega t
m a x i m u m space v e l o c i t y equals omega x subscript o
v equals negative omega x subscript o sin omega t
b u t square root of space sin squared theta plus c os squared theta end root equals 1 space left parenthesis P y t h. right parenthesis
omega r sin omega t equals omega square root of x subscript o squared minus x subscript o squared cos squared omega t end root equals omega square root of x subscript o squared minus x squared end root


Half an oscillation in stages:

  • The acceleration starts maximum and negative.
  • It reduces to zero in the centre.
  • Then the acceleration becomes positive, increasing to a maximum at the top on the left.

a equals negative a subscript 0 cos omega t equals negative omega squared x subscript o cos omega t
a equals negative omega squared x


If x = -xosinωt

a =

a is proportional to x but in the opposite direction.


Simple harmonic motion

Displacement and acceleration vary in the same way except that the acceleration is the opposite sign. We can say that the acceleration is proportional to the displacement but in the opposite direction:


This is the defining equation of simple harmonic motion. Note that this is only true if the swings of the pendulum are small. We will only consider examples where the motion is simple harmonic.

Defining quantities

Before we can derive a mathematical model we must define some terms and quantities.

Cycle: One complete swing from A to B to A.

Equilibrium position: The position where the pendulum bob would come to rest.

Amplitude (\(x_0\)): The maximum displacement from the equilibrium position.

Time period (\(T\)): The time for one cycle.

Frequency (\(f\)): The number of cycles per second (measured in Hertz, Hz).

\(f = {1\over T}\)

The equation for the sine function relating displacement and time is:

\(x=x_0\sin(2\pi f t)\)

You can see the effect of changing \(x_0\) and \(f\) using this simulation:

If the displacement varies according to the sine function then the velocity will vary according to cosine. The maximum velocity is given by the steepest gradient of the displacement vs time graph. In the following simulation you can see how this depends on both the frequency and maximum displacement, and that it in fact equals \(2πfx_0\):
So the equation for the variation of velocity with time is:
\(v=v_\text{max}\cos(2\pi f t)=2\pi fx_0\cos(2\pi ft)\)
The acceleration is the gradient of the velocity time graph which will be a -sin function. The maximum gradient of this curve is \(2πfv_0\) so the equation is:
\(a=-a_\text{max}\sin(2\pi ft)=-2\pi f\times 2\pi fx_0\cos(2\pi ft)=-(2\pi f)^2x_0\sin(2\pi ft)\)

So we can see that the magnitude of acceleration is proportional to displacement and the constant of proportionality is \((2πf)^2\):

\(a=-(2\pi f)^2x\)

\(\Rightarrow a\propto -x\)

Angular frequency (\(ω\)is equal to \(2\pi f\).

\(\Rightarrow a=-\omega^2x\)

If the frequency of an oscillating body is doubled but the amplitude stays the same, the maximum KE will change by a factor:

vmax = 2πfxo

KEmax = 1/2mvmax2


Analysing the forces

To understand why the motion is sinusoidal we need to have a look at the forces.

The motion is part of a circle so there must be a resultant force towards the centre, the centripetal force. This is greatest then the ball is moving fastest. At the bottom of the swing, you can see that the tension (\(C\)) is greater than the weight (\(mg\)). However, as we are only considering very small amplitude oscillations, the motion is almost horizontal. This enables us to make some approximations.

If the motion is horizontal there is no vertical acceleration. We can say that the vertical component of tension (now \(T\)) equals the weight:


The angle is very small so \(\cosθ = 1\):


Horizontally the forces are unbalanced so, according to Newton's 2nd law:


Using alternate angles and trigonometry, \(\sinθ={x\over L}\) so:

\(T{x\over L}=ma\)

\(T={maL\over x}\)

Equating the two equations for \(T\) gives, in terms of magnitudes:

\(mg={maL\over x}\)

\(a={g\over L}x\)

We can see that the magnitude of the acceleration is proportional to the displacement. The acceleration and displacement are also in opposite directions: 

\(a=-{g\over L}x\)

This means the motion is simple harmonic and so \(a=-\omega^2x\) applies. Equating acceleration in these two equations leads to an equation for the frequency of a simple pendulum:

\(\omega^2={g\over L}\)

\((2\pi f)^2={g\over L}\)

\(f={1\over 2\pi}\sqrt{g\over L}\)

Measuring acceleration due to gravity

We can see that the frequency of a pendulum depends upon acceleration due to gravity. We can use this to find \(g\) by measuring the frequency of a pendulum as we vary its length. A graph of \(f^2\) vs \(1\over L\) will be a straight line with gradient \(g\over 4π^2\).

Pendulum A has mass m and length L. Pendulum B has mass 2m and length 4L.

The ratio time period A / time period B =

T = 1/f so is proportional to √L


Energy changes

As the pendulum swings, energy is continually being exchanged between kinetic and potential. At the top of the swing the mass has only potential energy and at the bottom only kinetic energy. Assuming energy is conserved, the potential energy at the top will equal the kinetic energy at the bottom.

Potential energy

The potential energy at the top, \(E_\text{p}=mgh\) where \(h\) is the vertical height of the mass.

It's not obvious why that angle at the bottom is \(θ\over2\) so let's look at the geometry.

From the two triangles in the diagram we can deduce that if \(θ\) is small and measured in radians:

\(\theta={2h\over x}={L\over x}\)

\(h={x^2\over 2L}\)

If we take the lowest point to represent zero potential energy:

\(E_\text{p}=mgh={mg\over 2L}x^2\)

Since \(x\) varies sinusoidally with time, \(x=x_0\sin(\omega t)\), so potential energy at time \(t\) is:

\(E_\text{p}={mg\over2L}(x_0\sin(\omega t))^2={mg\over2L}{x_0}^2\sin^2(\omega t)\)

For a pendulum, \({g\over L}=\omega^2\), so:

\(E_\text{p}={1\over2}m\omega^2{x_0}^2\sin^2(\omega t)\)

Note that maximum potential energy (when the sine function is 1) is therefore:

\(E_\text{p, max}={mg{x_0}^2\over 2L}={1\over2}m\omega^2{x_0}^2\)

Kinetic energy

Kinetic energy is maximum at the bottom and zero at the top. If energy is conserved, the kinetic energy at the bottom will equal the potential energy at the top. Maximum kinetic energy is:

\(E_\text{k, max}={1\over2}m\omega^2{x_0}^2\)

We also know that kinetic energy is given by the formula \(E_\text{k} ={1\over 2}mv^2\), so maximum velocity is:

\(v_\text{max}=\omega x_0\)

If the displacement varies as a sine function then the velocity will be a cosine function:

\(v=v_\text{max}\cos(\omega t)=\omega x_0\cos(\omega t)\)

Kinetic energy at time \(t\) is:

\(E_\text{k}={1\over2}m\omega^2{x_0}^2\cos^2(\omega t)\)

Total energy

The total energy is constant and equal to the maximum potential energy (and kinetic energy!):

\(E_\text{T}={1\over 2}m\omega^2{x_0}^2\)

Mathematically we can see that, at any time \(t\):

\(E_\text{T}=E_\text{p}+E_\text{k}={1\over2} m\omega^2{x_0}^2\sin^2(\omega t)+{1\over2} m\omega^2{x_0}^2\cos^2(\omega t)={1\over 2}m\omega^2{x_0}^2(\cos^2(\omega t)+\sin^2(\omega t))\)

Using a mathematical identity (from Pythagoras): \(\cos^2(\omega t)+\sin^2(\omega t)=1\)

\(E_\text{T}={1\over 2}m\omega^2{x_0}^2\)


The amplitude of a real pendulum gets less with time due to the effect of air resistance.

This will reduce:

KE and PE are related, if one gets less so does the other.


When does the KE of a pendulum = PE?

It is half way in time but since velocity is not constant it is not halfway in terms of displacement

When KE = 1/2 KEmax

1/2mω2(xo2 - x2) = 1/4mω2xo2

2(xo2 - x2) = xo2

xo2 = 2x2

x = 1/√2 xo


A mass oscillates vertically on a spring.

At maximum displacement what can you deduce about the KE - gravitational PE - elastic PE?

The elastic PE will only be zero if the spring is at its equilibrium position.



Two oscillating bodies are said to be in phase if they are always travelling in the same direction.

A and B are in phase with each other but out of phase with C. The displacement vs time graphs for these oscillations are shown below.

We can see that the graph for C is the same as A and B except that it is shifted to the right by \(π\) radians, we say the phase difference is \(π\).

The phase difference between two oscillating bodies is defined by the phase angle. This is the angle by which you could shift the sine curve of one to be in phase with the other. In the simulation below you can see the effect of changing the phase angle:


The graph represents the displacement of two bodies oscillating with SHM.

red line:

The red line reaches the peak before the blue line so is in front.

You could also say that it lags behind the blue line by (2π -1) rads


SL Candidates

What you need to be able to do:

  • Define the quantities
  • Sketch the graphs for displacement, velocity and acceleration from different starting points
  • Understand the significance of the negative sign in \(a=-ω^2x\)
  • Sketch graphs of \(E_\text{p}\), \(E_\text{k}\) and total energy
  • Describe energy changes
Total Score:

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