1206
Text assignments:
General College Physics: Chapter 11, Sections 1-4. Do problems 7, 14, 24, 32, 33
University Physics: As on Assignment page for Chapter 9.
Recall from before that an equation of motion for a pendulum might be
x(t) = A cos( omega * t).
This is based on the solution to the equation x '' = - k x (University Physics) and on the circular model that results (University and General College Physics).
The associated velocity and acceleration functions are
v(t) = - omega A sin(omega * t) and
a(t) = -omega^2 A cos ( omega * t).
These functions can be reasoned out from the circular model, after observing that the speed and centripetal acceleration on the reference circle are omega A and omega^2 A.
They can also be derived as x ' (t) and x ' ' (t) by a simple application of the Chain Rule.
Experiment: First measure the horizontal velocity of a pendulum at the equilibrium position:
9 | 19 | 18 | |
12 | 26 | 25 | |
15 | 32.5 | 31 | |
18 | 39 | 39 | |
24 | 48 | 45 | |
27 | 58 | 44 | |
30 | 67 | 59 | |
33 | 74 | 60 | |
Predict the result that would be obtained for a pullback of 40 cm based on the SHM model, and compare with what we would get using PE change due to change in the vertical position of the pendulum.
How much work is required to pull a simple harmonic oscillator back from equilibrium to position x = A, assuming force constant k?
The force required to displace the oscillator is equal and opposite to the restoring force, so the force required to displace is F = k x.
Force therefore varies from k * 0 = 0 at the equilibrium position of the oscillator to k A at the x = A position.
So the average force is ( 0 + k A ) / 2 = 1/2 k A.
Ave force * displacement is 1/2 k A * A = 1/2 k A^2.
The restoring force constant for a pendulum is m g / L, so what should be its kinetic energy at equilbrium, assuming max displacement A?
The work required to displace to position x = A is 1/2 k A^2 = ( 1/2 m g / L) * A^2 = m g A^2 / ( 2 L). This energy is converted to KE as the pendulum comes back to equilibrium.
So what is the velocity of the pendulum at equilbrium?
KE = m g A^2 / ( 2 L) so
1/2 m vMax^2 = m g A^2 / ( 2 L) so
vMax = sqrt( g A^2 / L) = sqrt( g / L) * A.
What is this?
sqrt(g / L) = omega so this gives us vMax = omega * A.
Same thing we get from the circular model.
Figure out vMax, and get the equations of motion for x, v and a, for your pendulum.