- #1
Xyius
- 508
- 4
Homework Statement
A mass at the end of a spring with natural frequency [itex]\omega[/itex] is released from rest at position [itex]x_0[/itex]. The experiment is repeated byut now with the system immersed in a fluid that causes the motion to be overdamped with damping coefficient [itex]\gamma[/itex]. Find the ratio of the maximum speed in the former case to that in the latter. What is the ratio in the limit of strong damping? [itex]\gamma >>\omega[/itex] In the limit of critical damping?
2. The attempt at a solution
Right now I am in need with the first part as after I get this I think I should be fine with the other two parts. I just need to know if I am on the right track with my thought process.
So first I solve the undamped case (Which is frictionless).
[tex]-kx=x''m[/tex]
This leads to the solution..
[tex]x(t)=x_0 cos(\omega t)[/tex]
Taking the derivative and setting sine equal to 1, this has a max velocity of..
[tex]v_{max}=x_0 \omega[/tex]
This next part is the part I am unsure about. In the damped case, I solve the following differential equation..
[tex]x''m+bx'+kx=0[/tex]
And obtain a solution of..
[tex]x(t)=x_0 e^{-\gamma t}cos(\omega_0 t)+\frac{\gamma x_0}{\omega_0}e^{-\gamma t}sin(\omega_0 t)[/tex]
Where..
[tex]\gamma=\frac{b}{2m}[/tex]
and
[tex]\omega_0=\sqrt{\gamma^2+\omega^2}[/tex]
So my guess as to what I do next is to find the maximum of this velocity by differentiating x(t) twice and setting it equal to zero and getting a value of t. Then plug that value in and get a value of v. Doing all this I get a velocity of..
[tex]\frac{\gamma x_0 (\omega_0-\gamma)}{\sqrt{2\gamma^2 +\omega^2}}[/tex]
So the ratio would just be this value over the value I got in the frictionless case would it not? Something is telling me this isn't correct.
I apologize for not showing all my work, as it was a lot of work and would take forever for me to type up the code. If need by ill type it up in my Math Type program and post the image.
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