Comparing Maximum Speeds in Critically Damped and Undamped Harmonic Motion

AI Thread Summary
The discussion focuses on comparing the maximum speeds of a mass on a spring in undamped and critically damped harmonic motion. In the undamped case, the maximum speed is derived as v_max = ±x0*w, where x0 is the initial position and w is the angular frequency. For the critically damped scenario, the velocity function is more complex, leading to confusion in determining v_max. The key realization is that setting the velocity to zero does not yield the maximum speed, as it only indicates when the mass is at rest. Ultimately, the maximum speed in the undamped case is shown to be e times that in the critically damped case, resolving the initial confusion.
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Homework Statement



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

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?
 
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NATURE.M said:

Homework Statement



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

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?
Why can't you solve for ##v_\text{max}##? How do you find the maximum of a function?
 
vela said:
Why can't you solve for ##v_\text{max}##? How do you find the maximum of a function?

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.
 
Solving v(t)=0 would find when the velocity is 0, not when the velocity attains a maximum.
 
oh wow I feel pretty silly now. I can't believe I overlooked something so simple. Thanks a ton Vela.
 
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