# Finding Amplitude of spring oscillation after damping

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1. Dec 4, 2017

### AbigailG

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

A spring with spring constant 10.5 N/m hangs from the ceiling. A 520 g ball is attached to the spring and allowed to come to rest. It is then pulled down 6.20 cm and released.

What is the time constant if the ball's amplitude has decreased to 2.70 cm after 60.0 oscillations?

2. Relevant equations

I used:
T= 2π√m/k
A(t) = A(0)e^-λt

-I apologize for the notation, still learning how to post...

3. The attempt at a solution

Plugging into the equation for T I found the period to be 1.398 s
After 60 oscillations 83.895 seconds will have passed
This is the t I used in the amplitude function.

The initial amplitude is 0.0062 m
The final amplitude is 0.0027 m

0.0027 = 0.0062e^-λt
.43548 = e^-λt
ln(.43548)=-λt
ln(.43548)=-λ(83.895)
0.0099=λ

The time constant is 1/λ = 100.921

This is where I ended up and it is obviously wrong...haha. Any help would be greatly appreciated.

2. Dec 4, 2017

### kuruman

Why is it so obvious that it is wrong? What are the units for the time constant that you found? How does it compare with the time required for 60 oscillations?

On edit: 6.20 cm are 0.062 m and 2.70 cm are 0.027 m. Here it makes no difference because the erroneous factor is on both sides and cancels, however you should not live your life believing this lest you make an erroneous calculation where there is no cancellation.

3. Dec 4, 2017

### AbigailG

The units are seconds...i'm still not sure where I went wrong.
Without damping time < time with damping...Which seems to make sense because the spring is moving slower? Or would it take a longer amount of time for it to complete those oscillations...now I'm confused, I'm sorry.

I know it's wrong because when I enter it into Mastering Physics it's incorrect, haha.

4. Dec 4, 2017

### kuruman

The time constant is the time required for the amplitude to drop to $1/e$ of its value. Here, $6.20~cm/e = 2.28~cm$. If the time constant is 100 s you would expect the amplitude after 84 s to be greater that 2.28 cm so 2.70 cm is perfectly reasonable. I got the same answer as you.

What does Mastering Physics say is the correct answer? Mastering Physics has been sometimes discovered to be incorrect, hahaha.

5. Dec 4, 2017

### AbigailG

I wish I could tell you, but having a correct answer to work with would make things too easy, haha. I could click "give up" to find the answer but I don't want to lose the credit.

It seemed perfectly reasonable to me, it has to just be a simple mistake somewhere. Give it a shot if you're feeling ambitious! Haha, then we can compare, no pressure though. I appreciate your help thus far.

6. Dec 4, 2017

### AbigailG

Ahh, I see. That makes sense. So now all I would have to do is plug the new time (8.5 s) into my my final equation? 8.5 seconds seems much to quick to complete 60 oscillations of such a large pendulum...or am I understand the 8.5 seconds incorrectly? I'm sorry, this is really giving me trouble.

7. Dec 4, 2017

### RedDelicious

No. Completely disregard that post. I made an error when I punched in the numbers. I redid it and got the same time.

8. Dec 4, 2017

### kuruman

Don't put another solution then. Show your solution to your instructor, who has control over Mastering Physics, and ask him/her why it's wrong. If you're right (which I think you are) you'll get the credit regardless of what Mastering Physics says. This may be one of the occasions when Mastering Physics is in error.

9. Dec 4, 2017

### haruspex

Doesn't the damping affect the period?

10. Dec 4, 2017

### AbigailG

Okay I will. I have class tomorrow, I'll post an update tomorrow night. Thank you for your help!

11. Dec 4, 2017

### AbigailG

Yes, that is taken into account with the equation A(t) = A(0)e^-λt

As I understand it, the time used in the Amplitude function is the time taken without damping, the equation yields the amplitude with damping, and with the data we have we can solve for λ. But I could be wrong.

How would you take into account the increase in the period due to damping when solving for the time constant?

12. Dec 4, 2017

### haruspex

13. Dec 4, 2017

### AbigailG

Okay! I figured it out! The equation:

A(t) = A(0)e^-λt

was close...but not quite there, I looked further into my notes and found:

A(t) = A(0)e^-t/2tau

So...
0.0027 = 0.0062 e^-83.895/2(tau)
.43548 = e^-83.895/2(tau)
ln(.43548)=-(83.895/2tau)
tau = -83.895/2ln(.43548)
= 50.46 s

14. Dec 4, 2017

### kuruman

Hi @AbigailG, I am sorry for misleading you. Like you, I assumed that the time constant τ is just the inverse of λ forgetting the "doubling" effect of the negative amplitudes, i.e. that the "natural clock" for the decay is the half-period rather than the full period.

15. Dec 4, 2017

### haruspex

Fwiw, and in case this is not already clear, it turns out that although T= 2π√m/k is wrong here it is irrelevant.
The question is solved using the actual period, whatever it is. It is calculated from the 60 oscillation time and used directly to find the time constant.
(Specifically, the damped frequency is given by $\omega_1^2=m/k-\frac 14\nu^2$, where the kinetic equation is $m\ddot x+m\nu \dot x+kx=0$.)

16. Dec 6, 2017

### kuruman

My blind spot was not in the solution; it was in not realizing that, in terms of the time constant, the equation of motion can be written as $$\ddot{x}+\frac{1}{\tau}\dot{x}+\omega_0^2x=0.$$It's not a form that is normally seen for the damped harmonic oscillator, but it makes sense in retrospect.