Max Velocity of Damped/Undamped Spring

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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.

Homework Equations

The Attempt at a Solution

I think I have to get x(t) in each case, derive it twice, and then find the zeros. I’m not quite sure what x(t) should be in each case though. Would x(t) for the undamped spring maybe be Acos(wt)+Bsin(wt)? And then the max velocity (arctan(-a/b))/w?

 

[RedDelicious]

No that is wrong. You don't have to guess or try to memorize any formulas. If you aren't sure of the form of your solution, set up the equations of motion using Newton's second law and then solve the resulting ODE. That will give you your x(t). In both cases (damped and undamped), you get second order linear ODEs and so they are rather easily solved. Then just use the initial conditions given to solve for the coefficients.

To find the velocity, take the derivative of x(t). Then, if it's not immediately obvious, find the critical points and maximize it as you would any function.

In the second case, you will have to use the fact that the system is critically damped specifically when solving the ODE to get the correct form of x(t).

Maximize the velocity for this damped case in the same way and then take the absolute values of the maximum velocities to get the max speeds and compare them, which will show what you were supposed to.

 

[Fascheue]

No that is wrong. You don't have to guess or try to memorize any formulas. If you aren't sure of the form of your solution, set up the equations of motion using Newton's second law and then solve the resulting ODE. That will give you your x(t). In both cases (damped and undamped), you get second order linear ODEs and so they are rather easily solved. Then just use the initial conditions given to solve for the coefficients.

To find the velocity, take the derivative of x(t). Then, if it's not immediately obvious, find the critical points and maximize it as you would any function.

In the second case, you will have to use the fact that the system is critically damped specifically when solving the ODE to get the correct form of x(t).

Maximize the velocity for this damped case in the same way and then take the absolute values of the maximum velocities to get the max speeds and compare them, which will show what you were supposed to.

I see that I made a couple different mistakes, but shouldn’t x(t) still be Acos(wt)+Bsin(wt) in the first case? I would still have to solve for the coefficients using the initial conditions, but I would think that I could just use the equation for simple harmonic motion.

 

[RedDelicious]

I see that I made a couple different mistakes, but shouldn’t x(t) still be Acos(wt)+Bsin(wt) in the first case? I would still have to solve for the coefficients using the initial conditions, but I would think that I could just use the equation for simple harmonic motion.

Yep. The first case is just regular SHM.

 

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Yep. The first case is just regular SHM.

Okay good, I know what I did wrong there. Is there a formula that can be memorized for the critically damped oscillator? Or if not, what equation do I start with to get there?

 

[RedDelicious]

Okay good, I know what I did wrong there. Is there a formula that can be memorized for the critically damped oscillator? Or if not, what equation do I start with to get there?

No. You definitely do not want to try to memorize anything.

Start with Newton's second law, but this time, add in the damping force as well, which is proportional to and opposite in direction to the velocity. Post your equation if you're not sure if you've got it right.

 

[Fascheue]

No. You definitely do not want to try to memorize anything.

Start with Newton's second law, but this time, add in the damping force as well, which is proportional to and opposite in direction to the velocity. Post your equation if you're not sure if you've got it right.

Should I have not got the last x(t) from memorization?

Don’t I still need an equation first? If I were to have solved to get x(t) for the last part, I think I would have started with f=-kx, then applied Newton’s second law. Similarly do I not need an equation here?

 

[RedDelicious]

Should I have not got the last x(t) from memorization?

Don’t I still need an equation first? If I were to have solved to get x(t) for the last part, I think I would have started with f=-kx, then applied Newton’s second law. Similarly do I not need an equation here?

Yes that is the correct approach. You always start by finding the net force and then applying Newton's second law. In the first case, it was simply F = Fspring = -kx. For the second case, however, the spring force isn't the only force acting on it. You now have the damping force and so F = Fspring + Fdamp.

 

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Yes that is the correct approach. You always start by finding the net force and then applying Newton's second law. In the first case, it was simply F = Fspring = -kx. For the second case, however, the spring force isn't the only force acting on it. You now have the damping force and so F = Fspring + Fdamp.

Fspring is equal to -kx but what is Fdamp equal to?

 

[RedDelicious]

Fspring is equal to -kx but what is Fdamp equal to?

The damping force is directly proportional to the velocity, but works to slow it down and so its opposite in direction.

What does that give you?

 

[Fascheue]

The damping force is directly proportional to the velocity, but works to slow it down and so its opposite in direction.

What does that give you?

Would that give f = -kx - av, where a is some constant?

 

[RedDelicious]

Would that give f = -kx - av, where a is some constant?

Yep. Those are the only forces and so now you can apply Newton's second law and solve for x(t).

 

[Fascheue]

Yep. Those are the only forces and so now you can apply Newton's second law and solve for x(t).

Yep. Those are the only forces and so now you can apply Newton's second law and solve for x(t).

I plugged in x’’ = (-kx - ax’)/m into a differential equation solver and got the attached image as an answer. I’m not quite sure how to solve the differential equation myself but I don’t think we’re expected to be able to in my class. Does that look right? And if it does do I just find the values of A and B with the initial conditions?

 

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Does anybody know if vmax in the first situation is (w)(x0)sec(arcsin(-1)(tan(arcsin(-1))?

 

[RedDelicious]

I plugged in x’’ = (-kx - ax’)/m into a differential equation solver and got the attached image as an answer. I’m not quite sure how to solve the differential equation myself but I don’t think we’re expected to be able to in my class. Does that look right? And if it does do I just find the values of A and B with the initial conditions?

I can't imagine why you'd be assigned this problem if you weren't expected to be able to work with second order differential equations or learn how to at least. There should have been a primer on solving second order ODEs in the textbook or in class, even if you haven't taken a formal differential equations course yet.

No. That is the wrong form of the solution because you didn't take into account what it means for the system to be critically damped.

Check out this link on solving second order linear ODEs with constant coefficients. Lots of examples, including one that is similar to this problem.

http://www.stewartcalculus.com/data...ntexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf

Does anybody know if vmax in the first situation is (w)(x0)sec(arcsin(-1)(tan(arcsin(-1))?

That is wrong. Show your work so I can see what you're doing.

 

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I can't imagine why you'd be assigned this problem if you weren't expected to be able to work with second order differential equations or learn how to at least. There should have been a primer on solving second order ODEs in the textbook or in class, even if you haven't taken a formal differential equations course yet.

No. That is the wrong form of the solution because you didn't take into account what it means for the system to be critically damped.

Check out this link on solving second order linear ODEs with constant coefficients. Lots of examples, including one that is similar to this problem.

http://www.stewartcalculus.com/data/CALCULUS Concepts and Contexts/upfiles/3c3-2ndOrderLinearEqns_Stu.pdf
That is wrong. Show your work so I can see what you're doing.

I started with x(t) = Acos(wt)+Bsin(wt).

I then replaced x(t) with x0 and t with 0 because those are the initial conditions.

I got A = x0

I then took the derivative of the equation and got x’(t) = -x0wsin(wt) + Bwcos(wt).

Then I replaced x’(t) with 0 and t with 0 again because of initial conditions and got B = x0tan(wt).

That gave me x(t) = x0cos(wt)+x0tan(wt)sin(wt)

I then derived this twice to get the value of t when velocity is at its maximum, and then plugged that into my equation for v.

Is the equation x(t) = x0cos(wt)+x0tan(wt)sin(wt) and the process after correct? I feel like I most likely messed up the derivatives.

 

[RedDelicious]

I started with x(t) = Acos(wt)+Bsin(wt).

I then replaced x(t) with x0 and t with 0 because those are the initial conditions.

I got A = x0

I then took the derivative of the equation and got x’(t) = -x0wsin(wt) + Bwcos(wt).

Then I replaced x’(t) with 0 and t with 0 again because of initial conditions and got B = x0tan(wt).

That gave me x(t) = x0cos(wt)+x0tan(wt)sin(wt)

I then derived this twice to get the value of t when velocity is at its maximum, and then plugged that into my equation for v.

Is the equation x(t) = x0cos(wt)+x0tan(wt)sin(wt) and the process after correct? I feel like I most likely messed up the derivatives.

Where are you getting tan(wt) from? Your derivative is correct but your evaluation at t=0 is whacky.

[tex]x'(t)=-X_o\omega\sin{\omega t}+B\omega\cos{\omega t}
\\
x'(0)=0 => -X_0\omega\sin{0}+B\omega\cos{0}=0
[/tex]

Sin(0) = 0. Cos(0) = 1. What does this tell you about B?

 

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Where are you getting tan(wt) from? Your derivative is correct but your evaluation at t=0 is whacky.

[tex]x'(t)=-X_o\omega\sin{\omega t}+B\omega\cos{\omega t}
\\
x'(0)=0 => -X_0\omega\sin{0}+B\omega\cos{0}=0
[/tex]

Sin(0) = 0. Cos(0) = 1. What does this tell you about B?

Is B just zero then? And x(t) = x0cos(wt)?

 

[RedDelicious]

Is B just zero then? And x(t) = x0cos(wt)?

Yes. So now what is the max velocity and speed?

 

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Yes. So now what is the max velocity and speed?

I got t = 1/2pi for the time where v is at its maximum and the maximum value for velocity to be -wx0sin((w)(pi)(1/2))

 

[RedDelicious]

I got t = 1/2pi for the time where v is at its maximum and the maximum value for velocity to be -wx0sin((w)(pi)(1/2))

Not quite.

The w is part of the sine argument, and so when you're solving for t, there should be an omega in there.

 

[Fascheue]

Not quite.

The w is part of the sine argument, and so when you're solving for t, there should be an omega in there.

I did it again and got t = pi/(2w) and Vmax = -x0w.

 

[RedDelicious]

I did it again and got t = pi/(2w) and Vmax = -x0w.

Looks good. The max speed (not velocity) is then just x0w, the absolute value, which completes the first part of the problem.

You now just have to do the same thing for the second part with damping. Again, I highly recommend working through the link I shared on the second order equations. It will help you get the right form of your equation for x(t) when the system is critically damped.