Simple Harmonic Motion Waves 14.2 Simple Harmonic motion (SHM ) 14-3 Energy in the Simple Harmonic Oscillator 14-5 The Simple Pendulum 14-6 The Physical Pendulum and the Torsional Pendulum 15-1 Characteristics of Wave Motion 15-2 Types of Waves: Transverse and Longitudinal 15-3 Energy Transported by Waves FINAL STUDY GUIDE IS POSTED IN HW section

Example 14-7: Energy calculations. For the simple harmonic oscillation of Example 14–5 (where m=0.300kg, k = 19.6 N/m, A = m, x = -(0.100 m) cos 8.08t, and v = (0.808 m/s) sin 8.08t), determine (a) the total energy, (b) the kinetic and potential energies as a function of time, (c) the velocity at half amplitude (x = ± A/2), and (d) the kinetic and potential energies when the mass is m from equilibrium 14-3 Energy in the Simple Harmonic Oscillator

We already know that the potential energy of a spring is given by: The total mechanical energy is then: The total mechanical energy will be conserved, as we are assuming the system is frictionless Energy in the Simple Harmonic Oscillator

If the mass is at the limits of its motion, the energy is all potential. If the mass is at the equilibrium point, the energy is all kinetic. We know what the potential energy is at the turning points: 14-3 Energy in the Simple Harmonic Oscillator

The total energy is, therefore, And we can write: This can be solved for the velocity as a function of position: where 14-3 Energy in the Simple Harmonic Oscillator

This graph shows the potential energy function of a spring. The total energy is constant Energy in the Simple Harmonic Oscillator

14-5 The Simple Pendulum Example 14-9: Measuring g. A geologist uses a simple pendulum that has a length of cm and a frequency of Hz at a particular location on the Earth. What is the acceleration of gravity at this location?

14-5 The Simple Pendulum Problem (I) A pendulum has a period of 1.35 s on Earth. What is its period on Mars, where the acceleration of gravity is about 0.37 that on Earth?

A simple pendulum consists of a mass at the end of a lightweight cord. We assume that the cord does not stretch, and that its mass is negligible The Simple Pendulum

In order to be in SHM, the restoring force must be proportional to the negative of the displacement. Here we have: which is proportional to sin θ and not to θ itself. However, if the angle is small, sin θ ≈ θ The Simple Pendulum

Therefore, for small angles, we have: where The period and frequency are: 14-5 The Simple Pendulum

So, as long as the cord can be considered massless and the amplitude is small, the period does not depend on the mass The Simple Pendulum