Comparing Maximum Speeds in Critically Damped and Undamped Harmonic Motion

Click For Summary

Homework Help Overview

The discussion revolves around comparing the maximum speeds of a mass-spring system in two scenarios: one in simple harmonic motion and the other in critically damped motion. The original poster presents a problem from Morin's textbook, seeking to establish a relationship between the maximum speeds in these two cases.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive expressions for maximum speed in both cases, using trigonometric and exponential functions. Some participants question the method of finding maximum velocity, suggesting that setting the velocity to zero does not yield the maximum speed.

Discussion Status

Participants are engaged in exploring the problem, with some providing guidance on how to approach finding the maximum speed. There is an acknowledgment of confusion regarding the method used to determine the maximum speed in the critically damped case.

Contextual Notes

There is a noted difficulty in applying the correct method to find the maximum speed in the critically damped scenario, with assumptions about the behavior of the system at t=0 being questioned.

NATURE.M
Messages
298
Reaction score
0

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?
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 22 ·
Replies
22
Views
5K
Replies
16
Views
2K
Replies
9
Views
2K
Replies
5
Views
2K
Replies
3
Views
981
Replies
2
Views
1K
Replies
4
Views
1K
  • · Replies 14 ·
Replies
14
Views
3K
Replies
13
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K