April 198.8,9.1-9.3

April 219.1-9.3

April 23Boardwork

Quiz 9

April 2610.1-10.4

April 2810.5-10.6

April 30Boardwork

Quiz 10

May 311.1-11.4,

11.7

May 514.1-14.6

May 7Review

May 10 May 12Final Exam

May 14

Schedule

Final Exam: Wednesday, May 12, 10:30 am.

Chapter 11Vibrations and Waves

11.1 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.

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

*

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.

You studied Hooke’s Law in one of your labs. Hooke’s Law is valid for forces that take the form (in one dimension) Fx = - kx, where x is the displacement from equilibrium.

OSE: F= kx, restoring

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.

The displacement (x, or in this case, y) is measured away the equilibrium position.

The maximum displacement is called the amplitude A.

A complete back-and-forth (or whatever) motion is called a cycle.

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.

mv2/2 + kx2/2 = kA2/2 = mv02/2

11.2 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

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.

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.

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!

Click here to see this in action.

11.3 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.

For SHM, the frequency f is the number of complete cycles per second completed by the oscillator.

For circular motion, the angular velocity is the angle per second swept out by the radius vector, and is measured in radians per second.

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.

It is no coincidence that angular velocity and angular frequency have the same symbol and the same units.

There are 2 radians in a circle, so if we represent a cycle by one time around the circle, =2f, and there are 2 radians in a cycle. For SHM, the angular frequency also has units of radians per second.

x = A cos (t) = A cos (2f t)

v = -v0 sin (2f t) where v0 = 2Af

a = -a0 cos (2f t) where a0 = kA/m.

mT = 2π

k

1 kf =

2π m

2πT =

ω

Relationships between period, frequency, and angular frequency:

For SHM, the angular frequency isk

ω = 2πf =m

Using our analogy with circular motion:

All the equations on this page are OSE’s!

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?

The restoring force is F = mg sin.

11.4 The Simple Pendulum

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

.L

OSE : T = 2πg

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?

11.7 Wave Motion

Waves and springs demonstration.

Wave pulse traveling down string, striking “fixed” boundary.

Wave pulse traveling down string, striking “free” boundary.

Take a string, attach one end to an “arm” which can vibrate continuously up and down, and attach the other end to some rigid object.

When the vibrating arm moves up, a pulse is sent down the string:

vibrating arm attach string here

pulse

travels down string

reflects(& inverts)

meets another pulse on the way back

You end up with standing waves in the string. You may have done this experiment in your Phys 22 lab.

Things to note:

The wave pulse travels down the string. Does any matter travel down the string?

travels down string

The matter in the string moves up and then back down, but does not move down the string.

As we saw on the previous slide, if both ends of the string are fixed, a standing wave results.

If the string is “infinitely” long and the up-and-down pulses repeat, we get a traveling wave.

The traveling wave moves through the medium. Does any matter move along with the wave?

No!

The wavelength is defined as before.

The period is the time for the wave to travel one wavelength, and f = 1/T as before.