# Dampened Harmonic Motion

1. Oct 16, 2014

### NATURE.M

1. The problem statement, all variables and given/known data

A mass on the end of a spring is released from rest at position x0. The experiment is repeated, but now with the system immersed in a fluid that causes the motion to be critically damped. Show that the maximum speed of the mass in the first case is e times the maximum speed in the second case. This is question is out of Morin 4.28

3. The attempt at a solution

So for the initial case, we have x(t) = Acost(wt) => v(t) = -Awsin(wt). From initial conditions, we have x(0) = x0 and v(0) = 0. Using this, we find that A=x0. Thus v_max = +/- x0*w.

The next part is more confusing. Taking the solution to the critically damped case, we have x(t) = e^(-γt) * (A+Bt)
Then solving for the initial conditions, I have A = x0 and B = γ*x0. Taking the derivative of x, we have
v(t) = e^(γt) * (-γ^2 * x0 * t) . Note γ = w. But you can't solve for v_max from there. So I'm kinda stuck. Any advice?

2. Oct 16, 2014

### vela

Staff Emeritus
Why can't you solve for $v_\text{max}$? How do you find the maximum of a function?

3. Oct 17, 2014

### NATURE.M

Well in this case set v(t) equal to 0 and solve for t. So we have e^(γt) ≠ 0, and -γ^2 * x0 * t = 0 ⇒ t = 0
This makes no sense, since at t=0, you are at rest. This is why I'm stuck.

4. Oct 17, 2014

### vela

Staff Emeritus
Solving v(t)=0 would find when the velocity is 0, not when the velocity attains a maximum.

5. Oct 17, 2014

### NATURE.M

oh wow I feel pretty silly now. I can't believe I overlooked something so simple. Thanks a ton Vela.