1210

Note the review on the homepage under the Lectures button at the top of the page.

At the end of the Assignments page is a link to lecture notes from Fall 02.  The Class 1213 notes contain many problems from a practice final, starting about halfway down the page.  The material that precedes the practice final questions might be useful also.

A 117-cm pendulum starts out 10 cm from equilibrium, to the left of equilibrium, at t = 0.

It is given enough velocity that it swings out to a 15 cm displacement before beginning its return to equilibrium.

What speed did it have at the 10 cm position?

Listing things we know about this situation:

• Max displacement will be 15 cm.
• Mass is 1 kg.
• Length is 117 cm.
• g = 9.8 m/s^2.

Shoulda been listed first, before you even finished reading the problem:

• omega = sqrt(k/m)
• k = m g / L

Shoulda been listed promptly:

• PE = .5 k x^2

We are interested in something that has to do with the 10 cm and 15 cm positions.  It would be easy and reasonable to find the PE at both of these points.

We first note that k = m g / L = 1 kg * 9.8 m/s^2 / 1.17 m = 8.3 N / m, approx..

We obtain

• PE at 10 cm position = .042 J
• PE at 15 cm position = .094 J approx..

We assume that there are no dissipative forces.  In this case `dPE = - `dKE.

We conclude that from the 10 cm position to the 15 cm position we have `dKE = .042 J - .094 J = -.052 J.

We know also that the KE at the 15 cm position is 0.

From this we conclude that the KE at the 10 cm position musta been .052 J.  That's the only way it's gonna lose .052 J and end up with KE = 0.

We now easily find the velocity at the 10 cm position.  We have KE = .052 J so

• .5 m v^2 = KE, telling us that
• v = sqrt( 2 KE / m) = sqrt( 2 * .052 J / (1 kg) ) = .32 m/s.

Describe in detail the reference-circle model for this motion.

Angular frequency is omega = sqrt(k/m) = ... = 2.89 rad / sec, approx..

At t = 0 the reference circle position has the characteristic that the vertical line is at x = - 10 cm (this is 10 cm from equilbrium and to the left of equilibrium).

Our reference circle has radius 15 cm.  The reference circle position is therefore in either the second or third quadrant.

If the reference circle position is in the third quadrant the motion will start off to the right.

The angular position at the start is theta_0 = arccos(-10 / 15).

The equation of motion for the SHM is given by

• x = 15 cm cos(2.89 rad/s * t + theta_0).