Question

The ratio of successive amplitudes is A(n+1)=0.8A(n) The amount of energy stored in the oscillation is proportional to the square of the amplitude. Determine for these oscillations, the amount of energy lost as a percentage of the energy stored in the previous oscillation.
@experimentX

Answer 1

what oscillation?? harmonic oscillator?

Answer 2

Yes, harmonic oscillation.

Answer 3

This is the graph that's given.

Answer 4

I did it like this:
\[\frac{ E' }{ E } = \frac{ A ^{2} }{ (0.8A)^{2} }\]

Answer 5

So the energy of the next oscillation would be 0.64 times the energy of previous oscillation. Which means about 36% is lost. Am I right?

Answer 6

how do you model this recurrence relation?
A(n+1)=0.8A(n)

Answer 7

let your initial condition be A(0)

Answer 8

Actually, I calculated the ratio of the successive amplitudes.
A1/A2=A2/A3=A3/A4 and so on...

Answer 9

A(n) = 0.8^n A(0)

Answer 10

How do I do the energy calculation then?

Answer 11

hmm ... what do you mean by E ??

Answer 12

KE or ME?

Answer 13

Energy stored in the oscillation.

Answer 14

okay okay ...i presume it's ME then ... but BRB in 10 - 15 min

Answer 15

Okay.

Answer 16

you know that the equation of harmonic oscillator is given by
\[ y = A \sin( \omega t + \phi )\]
but A is dependent on ,,, time period.

Answer 17

Okay.

Answer 18

How does this relate with energy of oscillation?

Answer 19

calculate PE from y
calculate KE form y'

Answer 20

but you need to model \( A = A(t) \) such that \( A(n+1) = 0.8 A(n) \), we know that I comes back to same after time T, therefor it would be wise to model the system by using
\[ \huge y(t) = 0.8 ^{\frac{\omega t}{2\pi}} A(0) \sin( \omega t) \]

Answer 21

at some time \( \tau \) find the PE and KE. to find KE, you will need \( y'(x) \) so
\[ KE = \frac m 2 y'(x)^2\]

Answer 22

to find PE you need hooks law ... use
\[ PE = \frac 1 2 ky(x)^2 \]

Answer 23

find the total energy at time \( \tau \)
then find total energy at \( \tau + T \) and see what you get and what you an do.

Answer 24

Damn!! I only slept for 1 hrs in past 50 hrs.

Answer 25

I'm so sorry. But I think you are just complicating things.

Answer 26

what made things complicated?

Answer 27

See, its just a 2-mark question. The graph simply shows an object undergoing SHM that is underdamped. I calculated that the ratio of successive amplitudes stays constant i.e every successive amplitude is 0.8 times that of the previous amplitude.
The question states that energy is proportional to the square of amplitude, and asks to calculate how much energy would be lost in each successive amplitude.
Isn't it a simple ratio question?

Answer 28

of course ... if you know the Energy is proportional to square of amplitudes then why not do it yourself lol??

Answer 29

I asked you if I did it correctly or not.
I calculated the square of 0.8 which is 0.64. So I said that energy loss would be 36%. But the answer apparently is 64% loss.
Isn't 64% the energy stored in the next oscillation? Which would mean that 36% has been lost?

Answer 30

i think 64% should be stored as well ... check your question again.

Answer 31

So you mean that 36% would be lost, right?

Answer 32

yep ... of course.

Answer 33

Okay, thank you.

Answer 34

I am too sleepy right now ... i'll see this in detail tmr.

Answer 35

Okay. (: