Episode 302: Getting mathematical
Lesson for 16-19
- Activity time 120 minutes
- Level Advanced
The previous episode laid the qualitative foundations for what follows here: a mathematical approach to simple harmonic motion (SHM), starting from its physical basis and the forces involved.
- Demonstration and Discussion: The restoring force in SHM (20 minutes)
- Demonstration and Discussion: Graphical representations of SHM (30 minutes)
- Discussion: Equations of SHM (30 minutes)
- Discussion: Linking SHM to circular motion (10 minutes)
- Student activity: Making a computer model (30 minutes)
Discussion and demonstrations: The restoring force in SHM
So far, we have only considered the characteristics of SHM. Now we can go on to look at the underlying causes of this motion, in terms of forces.
Using a model system, look at the forces involved in SHM. Start with the trolley tethered by springs. Show that it remains stationary at the midpoint of its oscillation; it is in equilibrium. What resultant force acts on it? (Zero.) If it is displaced to the right, in which direction does it start to move? (To the left.) What causes this acceleration? (A resultant force to the left.) You can readily feel the force when you pull the trolley to one side.
So displacement to the right gives a resultant force (and hence acceleration) to the left, and vice versa. We call this force the restoring force (because it ties to restore the mass to its equilibrium position). The greater the displacement, the greater the restoring force.
We can write this mathematically:
Restoring force F ∝ displacement x
Since the force is always directed towards the equilibrium position, we can say:
F ∝ -x
F = − k × x
Where the minus sign indicates that force and displacement are in opposite directions, and k is a constant (a characteristic of the system).
This is the necessary condition for SHM.
Now it should be clear that we are dealing with vector quantities here; what are they? (Displacement, velocity, acceleration, force.) It makes sense to use a sign convention. We call the midpoint zero; any quantity directed to the right is positive, to the left is negative. (For a vertical oscillation, upwards is positive.)
Think about mass-spring systems. Why might we expect a restoring force which is proportional to displacement? (This is a consequence of Hooke’s law.)
It is harder to see this for a simple pendulum. As the pendulum is displaced to the side, the component of gravity restoring it towards the vertical increases. Force and displacement are (roughly) proportional, and proportionality is closest at small angles. For this reason, it will make sense to start a mathematical analysis with mass-spring systems.
Discussion and demonstrations: Graphical representations of SHM
We need to get to a point where we can develop the equation F = − k × x to a = − ω 2 × x , where a is the acceleration and ω is the angular velocity associated with the SHM. To do this, we develop the graphical representation of SHM.
Consider first the tethered trolley at its maximum displacement. Its velocity is zero; as you release it, its acceleration is maximum. Show how the trolley accelerates towards the midpoint. Sketch the displacement-time graph for this quarter of the oscillation. What happens next? The trolley decelerates as it moves to maximum negative displacement. Extend the graph. Continue for the second half of the oscillation, so that you have one cycle of a sine graph.
Below this graph, draw the corresponding velocity-time graph. Deduce velocity from the gradient of the displacement graph. Show how maximum velocity occurs when displacement is zero.
Below this, sketch the acceleration-time graph.
Discussion: Equations of SHM
These graphs can be represented by equations. For displacement:
x = A × sin(2 π ft) or x = A × sin( ω t)
Explain that f is the frequency of the oscillation, and is related to the period T by f = 1 T . The amplitude of the oscillation is A .
Velocity: v = 2 π ft × A × sin(2 π ft) or v = ω × A × sin( ω t)
Acceleration: a = (2 π ft)2 × A × sin(2 π ft) or v = ω 2 × A × sin( ω t)
Depending on your students’ mathematical knowledge, you may be able to explain where these equations come from. You may simply have to indicate their plausibility.
Also point out that in some text books the sin and cos functions are written in terms of the
real physical frequency f rather than ω.
Comparing the equations for displacement and acceleration gives:
a = − ω 2 × x
and applying Newton’s second law gives:
a = − m × ω 2 × x
These are the fundamental conditions which must be met if a mass is to oscillate with SHM. If, for any system, we can show that F ∝ – x then we have shown that it will execute SHM, and its frequency will be given by:
ω = 2 π f , so ω = F mx therefore ω is related to the restoring force per unit mass per unit displacement.
Discussion: Linking SHM to circular motion
If your specification requires you to explore SHM with reference to
motion in the auxiliary circle (or if you wish to adopt this approach anyway), then this is a good point to do so. It has the merit of showing the link between SHM and circular motion, but for many students it may simply add confusion.
See also this tutorial on the University of Guelph website,which links ω for a circle and SHM.
Student activity: Making a computer model
It will help students to grasp the relationships between displacement, velocity and acceleration in SHM if they make a computer model.