Evans SPH 4U1

Physics Grade 12

Unit 5: Waves

Note 1: Simple Harmonic Motion


Reference: Chapter 4.5


We begin our wave unit with a discussion on Simple Harmonic Motion since waves are an application of Simple Harmonic Motion.


Simple Harmonic Motion

When any motion repeats itself, back and forth, vibrating over the same path, the motion is described as periodic (an example is a spring and a mass). If a mass attached to a spring is pushed towards the spring (compressing it and storing energy in it) then released, the mass attached to the spring oscillates back and forth about the equilibrium position. When the mass stretches or compresses the spring, the spring exerts a force on the mass that acts to "restore" the spring to its equilibrium position. If there is no friction, the system will oscillate forever. Hooke's law describes this restoring force, or F = -kx. Any vibrating system for which the restoring force is directly proportional to the negative value of the displacement x (or one that obeys Hooke's law) is said to be in Simple Harmonic Motion (SHM).

Simple Harmonic Motion Terms::

Displacement The distance x of the object from the equilibrium position

Amplitude The maximum displacement from the equilibrium position.

Period (T) The time it takes for one cycle (or back and forth motion).

Frequency (f) The number of cycles completed per second (SI unit is Hz, or Hertz).

Energy in the Simple Harmonic Oscillator:

Work is done when a spring is stretched or compressed. The elastic potential energy of the spring is given by

Ee = 1/2 kx2
Thus the total mechanical energy (ET) is the sum of the kinetic and potential energies at that point.
ET= 1/2 kx2 + 1/2 mv2

At its maximum amplitude  there is no Ek so ET =Ee
    ET = Ee = 1/2 kx2 or Ee = 1/2 kA2 where A is the amplitude

At the equilibrium point the maximum velocity (vo) occurs.  At this point, there is no Ee so  ET =Ek (maximum kinetic energy equals the total mechanical energy of the oscillator).
    ET =Ek= 1/2 mvo2

An object in SHM has its greatest acceleration at its greatest amplitude and its maximum velocity at its equilibrium point. The acceleration is given by

Period of a Simple Pendulum:


Period of SHM The period of an object in SHM is dependent upon the stiffness of the spring (related to k) and the mass (m) that is oscillating. The period does not depend upon the amplitude.  

                Period of SHM formula
A simple harmonic oscillator obeys Hooke's Law, F=kx. If a mass is hung on a spring and the spring is allowed to come to rest, the spring constant can be calculated knowing the mass and the displacement, x. If the mass is then pulled down and allowed to oscillate back and forth about the equilbrium point, the period of oscillation can be found using the above formula.

Example 1:

An object follows SHM. If the acceleration at any time is given by a = 4x where x is the position, what is the period? (Answer: 3.14)


Example 2:

A mass-spring system undergoing SHM has a maximum energy of 6.00 J. The mass is 0.15kg and the force constant is 250 N/m.

a) What is the amplitude of the vibration? (Answer: 0.22 m)

b) Determine the maximum speed of the mass. (Answer: 8.94 m/s)

c) Calculate the speed of the mass when it is 10.0 cm from the equilibrium position. (Answer: 7.96 m/s)


Example 3:

Two springs with the same spring constant k = 39.5 N/m are connected to two different masses 1 kg and 4 kg. Both springs are pulled 4 cm below the equilibrium point and released at the same time and allowed to oscillate for 4.8 s. How many times were the masses at the same position at the same instant in time? (Not counting the start). Assume that the equilibrium point is the same for both systems. (Answer: 7)


Text Questions:

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