Simple Harmonic Motion
What is SHM (Simple Harmonic Motion)
- An object moving with SHM oscillates to and fro, either side of an equilibrium position.
- This equilibrium position is the midpoint of the objects motion.
- This distance of the object from the equilibrium is called the displacement.
- There is always a restoring force pulling or pushing the object back towards its equilibrium position.
- The size of the restoring force depends on the displacement.
- The restoring force makes the object accelerate towards the equilibrium.
Definition of SHM
- An oscillation in which the acceleration of an object is directly proportional to its displacement from the equilibrium position and directed towards the equilibrium.
-
\[\mathbf{a} \propto -\mathbf{x}\]
SHM on a graph
- Displacement is measured as a sine or cosine wave. It has a maximum value of A.
- Velocity is the gradient/derivate of the displacement graph. It has a maximum value of \(\omega A\)
- Acceleration is the 2nd derivate of the displacement graph. It has a maximum value of \(\omega^2A\)
A 50kg mass on the end of a spring is undergoing SHM. The potential energy of the mass is 28J at the maximum displacement. Calculate the maximum velocity that the mass reaches while oscillating.
- \(28=\frac{1}{2}\times50\times v^2\)
- \(28=25\times v^2\)
- \(\sqrt{ \frac{28}{25} }\)
- \(1.058\)
- \(1.1\) to 2
A girl with a mass of 35 kg is sitting on a swing, which is undergoing simple harmonic motion. To start the motion, she raised the swing 0.40 m above its lowest position, then lifted her feet off the round.
a) Find the maximum velocity the girl reaches on the swing.
- \(\frac{1}{2}mv^2=mgh\)
- \(mgh=35\times9.81\times 0.40=137.34\)
- \(\frac{1}{2}mv^2=137.34\)
- \(mv^2=274.68\)
- \(v=2.8014\)
- \(=2.8\) to 2 s.f
b) Sketch a graph of the girl's kinetic energy against time for one complete oscillation of the swing.
A pendulum is undergoing simple harmonic motion. The maximum displacement of the pendulum is 0.60 m and the maximum velocity of the pendulum is 0.90 ms\(^{-1}\).
a) Frequency?
- \(2\pi fA=0.9\)
- \(A=0.6\)
- \(2\pi f=1.5\)
- \(f=\frac{1.5}{2\pi}=0.24Hz\)
b) Time taken to complete one cycle: Time period
- \(T=\frac{1}{f}\)
- \(\frac{1}{0.239}=4.189\) seconds
c) Maximum acceleration = \(\omega^2A\)
\((2\pi f)^2A\)
\((2\times \pi \times 0.239)^2\times 0.6=1.353\) ms\(^{-1}\)