Simple Harmonic Motion – The maths stuff

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Simple Harmonic Motion – The maths stuff 4.1.3 Define simple harmonic motion (SHM) and state the defining equation as a =−ω2x . 4.1.4 Solve problems using the defining equation for SHM. 4.1.5 Apply the equations: v=v0 sin ωt , v=v0 cos ωt, v=±ω(x02− x2) , x=x0 cos ωt x=x0 sin ωt as solutions to the defining equation for SHM. 4.1.6 Solve problems, both graphically and by calculation, for acceleration, velocity and displacement during SHM.

1 Students are expected to understand the significance of the negative sign in the equation and to recall the connection between ! and T.

To do Carry out the experimental circus Try out some other oscillating systems; are they isochronic? Do they appear to be SHM? Students in pairs try out a selection and report back to a plenary session.

Further information Text p108 http://hyperphysics.phy- astr.gsu.edu/hbase/HFrame.html

Building a mathematical description Using a model system, look at the forces involved in SHM. Remember the trolley tethered by springs, 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. 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.

The restoring force in SHM 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 or F = - kx 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. 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.

Graphical representations of SHM Consider first the tethered trolley at its maximum displacement. Its velocity is zero; as you release it, its acceleration is maximum. 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.

T is the time period We describe one complete cycle as changing through an angle of 360o or 2π radians Angular velocity is Frequency = 1/T

Equations of SHM These graphs can be represented by equations. For displacement: x = A cos 2πft or x = A cos wt 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 Velocity: recognise and use v = v0cosωt, v = +/-ω(x02 – x2). In some text books the sin and cos functions are written in terms of the ‘real physical frequency’ f rather than w.

Acceleration and Force Acceleration: a = w2 A sin wt Comparing the equations for displacement and acceleration gives: a = - w2x a=+/-(2πf)2x and applying Newton’s second law gives: F = - m w2x

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: w = 2πf = √ (F/mx) so w is related to the restoring force per unit mass per unit displacement.

To do Read p 110-111 Answer all questions Homework – read and make notes on p 112-113 Answer questions 3,5 & 6