damped oscillations

**quietrain** · Nov8-09, 09:49 AM

the formula for damped oscillations is given as x = Ae^(-bt/2m) cos(ωt+Φ)

so why is the amplitude as a function of time given as only the first part?

meaning only A(t) = Ae^(-bt/2m)

it "ignores" the 2nd term which is the oscillating cosine term. which still encompass a time t value too...

so how come the amplitude function is given as so ?

thanks

**quietrain** · Nov9-09, 03:28 AM

anyone knows?

**MikeyW** · Nov9-09, 06:01 AM

Are you asking why the amplitude decay is independent of the wave shape?

A wave is in the form x = A*cos(wt+phi), there is no reason that A, w, phi cannot also vary with time to give a more complicated wave. The multiplication of A with the wave shape (the cosine term) means the wave gets smaller in height as time goes on, which reflects our observation of decaying oscillations.

I don't really understand the question,
Do you disagree that the function represents a decaying wave?

**quietrain** · Nov9-09, 08:10 AM

oh isee.. thanks

so you meant that the term Ae^-bt/2m signifies the decaying amplitude magnitude

whereas the cosine term represents the shape of the wave.

so if there were no damping, the A term would just be A without the epsilon? hence no decaying and it is just the original amplitude.

right?

thanks

**MikeyW** · Nov9-09, 08:27 AM

Yes, exactly (although it is not an epsilon, its an exponential).

The graph below shows the final wave, and the amplitude is the dotted line. So the overall wave is being "squashed" over time because the amplitude is dropping.

**quietrain** · Nov10-09, 03:03 AM

oh ya... its exponential lol

er from your graph, does it always mean that within the same section of 1 wavelength, like the 2nd crest to the 3rd crest,

the amplitude( from middle to 2nd crest , and from middle to 2nd trough), is the same?

**MikeyW** · Nov10-09, 04:37 AM

No, the amplitude is continuously falling over all time.

**quietrain** · Nov10-09, 09:15 AM

ah ic ... thanks

**Bob S** · Nov10-09, 07:07 PM

Use exponentials to represent the signal:

x = Ae^-bt/2m(e^iwt+ e^-iwt)/2

The attenuation represents the real part of the function which has both real and imaginary parts, the imaginary being the oscillation.
Bob S

**quietrain** · Nov10-09, 09:33 PM

oh my. thats too complicated at the moment, i haven't really touched complex numbers yet

thanks though