Dampened Harmonic Motion

1. Apr 19, 2008

seichan

[SOLVED] Dampened Harmonic Motion

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

A mass M is suspended from a spring and oscillates with a period of 0.900 s. Each complete oscillation results in an amplitude reduction of a factor of 0.985 due to a small velocity dependent frictional effect. Calculate the time it takes for the total energy of the oscillator to decrease to 50 percent of its initial value. HINT: The amplitude after N oscillations=(initial amplitude)x(factor)^N.

2. Relevant equations
Newton's Second Law dampened- F=-kx-bv
x(t)=Ae^(-bt/2m)cos((dw/dt)t)
Energy=1/2kA^2 OR 1/2kv^2 [NOT constant]

3. The attempt at a solution
Alright, I am having a difficult time setting this problem up. I do not know how to use the hint either... So far, I have set it up this way:
F=-kx-bv
The integral of this is equal to work. Negative work is equal to KE.
-1/2kx^2+1/2bv^2=1/2kv^2
I realized that this is as far as I can get with this, so I tried to use the x(t) equation. However, we are not given a value for the mass. I am very frustrated right now so any direction you can give would be very much appreciated.

2. Apr 19, 2008

alphysicist

Hi Seichan,

I believe the correct approach is to use the total energy is (1/2)kA^2. You know how much the amplitude decreases with each oscillation, you know how long each oscillation takes, and you know you need the total energy to be 50%. Try that and see if it works for you.

3. Apr 19, 2008

seichan

Thank you. I do not where to go with that, though, because of having no initial Amplitude. I have, however, derived that I am looking for t=ln(2)/(2*angular acceleration). However, I do not know how to get the angular acceleration without having a known mass value. (a. acceleration=b/2m) Any ideas? Thanks again so much.

4. Apr 19, 2008

alphysicist

You do not need the initial amplitude, because you do not need the initial energy. All they ask is for the energy to decrease by 50%. Call the initial amplitude $A_0$. Now they want the energy to decrease by 50%. Write an expression for the initial energy, write an expression for the final energy, and then relate them. What do you get?

5. Apr 19, 2008

seichan

(.5k(A*.985^n))/2=.5kA
.5(A*.985^n)=A
A*.985^n=2A
.985^n=A

I feel like I'm missing something there, due to the fact that answer makes no sense. Thanks for baring with me.

6. Apr 19, 2008

alphysicist

I think you have a few mistakes here.

Initial energy is 0.5 k A^2

final energy is 0.5 k (.985^N A)^2

final energy = 0.5 initial energy

(Also in going from the third to the fourth line of your post the A's would have cancelled, which is why you don't need them.)

7. Apr 19, 2008

seichan

Alright. I got you there. My new arithmatic looks like this:

.5kA^2=[.5k(.985^n*A)^2]/2
kA^2=.5k(.985^n*A)^2
2A^2=(.985^n*A)(.985^n*A)
2=(.985^n)^2
ln(2)=.985^n

Ack! I'm truly sorry- it's been a long night and barely anything seems coherent anymore...

8. Apr 19, 2008

alphysicist

That's close, but there's two last issues here. The final energy is 1/2 the initial energy, so you need the first line to be

(1/2)* .5kA^2=[.5k(.985^n*A)^2]

Working down, we get a line like your fourth line, which is (doing your steps but starting with the above):

(1/2) = 0.985 ^(2n)

When you take the natural log, the exponent comes down:

ln(1/2) = 2n ln(0.985)

9. Apr 19, 2008

seichan

Thank you so much again. That was correct =) Your help was greatly appreciated!

10. Apr 19, 2008