# Driven, Damped Oscillations

## Homework Statement

I have a mass, B, attached to a vibrating wall. The wall is vibrating at two frequencies, .01 Hz and 75 Hz (later in the problem it turns out that we want to transmit the .01 Hz but not the 75 Hz oscillations). The spring constant is k=154 N/m, the mass of the block B is 54.7 kg, and the damping constant, b, is .765. The wall vibrates with amplitude A=0.05 m.

1) Explain why the expression x(t) which represents the position of B as a function of time is simply A(cos(ω1t)) added to A(cos(ω2t)).

2) If you wait until steady state, what is the ratio of amplitude of unwanted oscillations (at 75 Hz) to the amplitude of wanted oscillations (at .01 Hz)?

3) Now you cut the spring in half and attach a mass in the middle which experiences a damping constant of b'=.165. Compare the force exerted on the box B at 75 Hz for this new configuration to the old configuration.

## The Attempt at a Solution

I have written the differential equation of this as a driven, damped oscillator (driven because the wall is moving back and forth).

I end up with the equation: m x''(t) + b x'(t) + k x(t) = F0(cos(ω1t)+cos(ω2t)) simply by applying Newton's law... I'm pretty sure this is right so far.

But I'm not sure how to solve this! For part (1), it says explain-- and it makes sense that this is essentially like two different oscillators, each at its own frequency (since the cosines are linearly independent with the different frequencies) and so I could presumably just add up the two independent solutions. But surely this is not rigorous enough. So how do I solve the differential equation with the sum of cosines to show that x(t) really is the sum that we want?

For part (2), I'm thinking I would treat this as a normal driven/damped oscillator, and therefore I could just say that Amplitude = (k A / m) / sqrt[(ω022)2+(γω)2]... evaluate this for each frequency, and find the ratio. But I get something like 3E-5, which seems quite small.

For part (3), I haven't started quantitatively, but I think I would treat the NEW mass like a driven/damped oscillator while the block B is now not experiencing a driving force, but simply responds to the spring force brought on by the driven mass. I don't know what to do from here, though! (Also, since the spring was cut in half does each half now have double the spring constant-- we learned that this would happen if you cut a spring in half, but I feel like we immediately put them back in series so how could k just double?)

THANKS!

## Answers and Replies

lightgrav
Homework Helper
1) if you want to prove a solution, just take the derivatives and see if it works (brute Force). ("explain" likely means "describe in words what happens when you do that")
2) I got resonance at omega =5/3 , which is far from 12 (but even farther from .0016). it's only the terrible damping that has either of them above the tails.
3) EACH piece of spring has 2x the stiffness ... BOTH spring halves now try to RESTORE equilibrium. so the effective k becomes 4x what the original was.
... but the damping is now only 1/4 of what it used to be, so it might actually not be better! (10/3 is still far from 12)