Dampened Harmonic Motion and oscillation

[seichan]

[SOLVED] Dampened Harmonic Motion

Homework Statement

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.

Homework 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]

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.

 

[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.

 

[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.

 

[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?

 

[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.

 

[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.)

 

[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...

 

[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)

 

[seichan]

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

 

[alphysicist]

Great! Glad I could help.