Oscillation damped oscillations ? how to calculate energy after t

[masterburn]

Homework Statement

3. A damped oscillator's amplitude dec¡eases from 8 cm to 4 cm in 20 seconds, If the intial energy of the oscillator is 64 J, what is the energy âfter 40 seconds? (Recall: E: (l/2)kA2)

Homework Equations

not sure how to approach the problem

The Attempt at a Solution

intia E which is 64J = k*A^2 solve for k since 64 = K * 0.08^2. k = 9375 N

problem how do I take the damped oscillator into account for after 40 seconds

so after 40 seconds E = 9375*A^2 ? or am i doing it wrong totally

 

[ehild]

Do you know the mathematical expression as the amplitude of a damped oscillator changes with time? ehild

 

[masterburn]

is it A=Ae^-t/2T where T= M/b ?? I'm not sure what the time constant is in this case ? or would i apply the formula A = sqrt( xo^2 + (vo/w)^2) ?? in which case i have no idea what the v or the w and the xo is since distance is constantly changing

 

[ehild]

Read the problem: The amplitude decreases from 8 cm to 4 cm in 20 seconds. The amplitude decreases exponentially, as you have written, A=A_initiale^(-t/2T). Plug in the data: what is the time constant?

ehild

 

[masterburn]

would the time constant be 20 seconds since amplitude is decreased at 20 seconds, since T = m/b how do i calculate for T if the time constant isn't 20 seconds, but I think the time constant is indeed 20 seconds since the amplitude changes at 20 seconds

 

[ehild]

The amplitude decreased to half in 20 seconds. The energy is proportional to the amplitude. So the energy would decrease to 1/4 of the initial value in 20 seconds. In 40 seconds, the amplitude halves again. What about energy?

If general case, when the amplitude changes from A1 to A2 in t seconds,

$$A_2=A_1 e^{-\frac{t}{2T}}$$

$$\frac{t}{2T}=\ln(\frac{A_1}{A_2 })$$

ehild

 

[masterburn]

ahh i see so using the ln equation i can use the ln equation using A1 = 8 cm and A2= 4 cm, then I can figure out T ?

then using that constant I can figure out A = (0.08m)*e^(-t/2T) ? with that I can plug into E= KA^s ?

 

[masterburn]

but would the K apply since i found K using intial energy and initial amplitutde

 

[ehild]

ahh i see so using the ln equation i can use the ln equation using A1 = 8 cm and A2= 4 cm, then I can figure out T ?

Yes.

then using that constant I can figure out A = (0.08m)*e^(-t/2T) ? with that I can plug into E= KA^s ?

Yes, but you can use the energy directly. As the energy is proportional to the square of the amplitude, the energy also decreases exponentially
$$A_2^2=A_1^2(e^{-\frac{t}{2T}})^2=A_1^2e^{-\frac{t}{T}}$$
$$E_2=E_1e^{-t/T}$$.

E1=64J, and you know T.


ehild

 

[masterburn]

so the k will still work plugging into second E=K*A^2 with the new A, and thank you did not know most of the equations you posted my prof sucks! got all the hard equations

 

[ehild]

I am glad if it cleared ...

ehild