Simple Harmonic Motion

1. Simple Harmonic Motion A restoring force is one that moves a system back to an equilibrium position. Example: mass on frictionless table, attached to spring.

2. Example: gravity acting on a mass hanging from a string. * Example: gravity acting on a mass hanging from a spring. *Lots of material borrowed from: http://hyperphysics.phy-astr.gsu.edu/hbase/permot.html#permot

3. When the restoring force is linearly proportional to the amount of the displacement from equilibrium, the force is said to be a Hooke’s Law force. Physicists are always sloppy with their signs when doing Hooke’s law calculations.

4. OSE: F= kx, restoring We have studied Hooke’s Law already. Hooke’s Law is valid for forces that take the form (in one dimension) Fx = - kx, where x is the displacement from equilibrium. It is easy to show with calculus that, in the limit of small displacement, all restoring forces are Hooke’s Law forces. Systems consisting of Hooke’s Law forces undergo simple harmonic motion when displaced away from equilibrium. Such systems are often called simple harmonic oscillators. Examples of simple harmonic motion: mass bouncing on spring, swinging pendulum.

5. Here’s that calculus I mentioned. Perhaps you recall from math how functions can be written in the form of a Maclaurin’s series (a Taylor series about the origin): If F represents a restoring force (a force that “pulls the system back to the origin”) then F(0) = 0. For small displacements x, all the higher order terms (involving x2, x3, etc.) are small, so The – sign enters because F is a restoring force, so the derivative is negative.

6. The displacement (x, or in this case, y) is measured awaythe equilibrium position. The maximum displacement is called the amplitude A. A complete back-and-forth (or whatever) motion is called a cycle.

7. The time for one cycle is the period, T. The number of cycles per second is the frequency. We’ll learn about angular frequency shortly.

8. Energy in the Simple Harmonic Oscillator We get our OSE for the total mechanical energy of a simple harmonic oscillator by considering a mass attached to a spring on a horizontal, frictionless surface: E = mv2/2 + kx2/2 (here I am using x instead of s). At the equilibrium position (x=0), the energy is all kinetic. At x=A the energy is all potential. Thus mv2/2 + kx2/2 = kA2/2 = mv02/2 where v0 is the speed at x=0 where the energy is all kinetic. Your text solves this for v(t) and makes it look like a brand new equation, when it is really just old stuff.

9. Example 11-3. A spring stretches 0.15 m when a 0.3 kg mass is hung from it. The spring is then stretched an additional 0.1 m from this equilibrium point and released. Determine (a) the spring constant k, (b) the amplitude of oscillation A, (c) the maximum velocity v0, (d) the magnitude of the velocity when the mass is 0.04 m from equilibrium, and (e) the magnitude of the maximum acceleration of the mass.

10. Wait a minute! Where did the gravitational potential energy go? Once you confirm that a system undergoes simple harmonic motion, all the equations for a simple harmonic oscillator apply. The total mechanical energy of the system is the kinetic energy of the object oscillating, plus the potential energy associated with the restoring force. You need not worry about other details of the system! This is abstract, but powerful!

11. The Periodic and Sinusoidal Nature of SHM Important: the force in SHM is not constant, so the acceleration is not constant, so you can’t use the equations of kinematics. An object rotating in a circle (which is has a frequency and a period) is mathematically analogous to the SHM of a vibrating spring. Click here to see this in action.

12. Your text shows that T=2(m/k)1/2 and f=(1/2)(k/m)1/2. Before, for an object undergoing circular motion,  was the angular velocity and T=2/ the time for a revolution. For circular motion, the angular velocity  is the angle per second swept out by the radius vector, and is measured in radians per second. For SHM, the frequency f is the number of complete cycles per second completed by the oscillator.

13. There are 2 radians in a circle, so if we represent a cycle by one time around the circle, =2f, and there are 2 radians in a cycle. For SHM, the angular frequency  also has units of radians per second. It is no coincidence that angular velocity and angular frequency have the same symbol and the same units.

14. For SHM, the angular frequency is Relationships between period, frequency, and angular frequency: Using our analogy with circular motion: x = A cos (t) = A cos (2f t) v = -v0 sin (2f t) where v0 = 2Af a = -a0 cos (2f t) where a0 = kA/m. All the equations on this page are OSE’s!

15. Example 11-7. The cone of a loudspeaker vibrates in SHM at a frequency of 262 Hz. The amplitude of the center of the cone is A=1.5x10-4 m, and at t=0, x=A. (a) What is the equation describing the motion of the center of the cone. (b) What is the maximum velocity and maximum acceleration? (c) What is the position of the cone at t=1x10-3 s?

16. The Simple Pendulum The restoring force is F = mg sin. In the limit of small , this becomes F = mg  (restoring). This is a Hooke’s Law force and SHM occurs. Your text shows that the period is

17. For SHM, the period does not depend on the amplitude. For a pendulum with small , this is true, so a pendulum exhibits SHM for small displacements. For large  (greater than 15 degrees or so) the small-angle approximation is not valid and the period does depend on the amplitude (max). Example 11-8. (a) Estimate the length of the pendulum in a grandfather clock that ticks once per second. (b) What would be the period of a clock with a 1 meter long pendulum?