Energy lost in a lightly damped osciallator

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Homework Help Overview

The discussion revolves around the energy loss in a lightly damped oscillator, specifically focusing on how the amplitude decrease affects the energy over cycles of oscillation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between amplitude and energy in simple harmonic motion (SHM) and how energy changes with a decrease in amplitude due to damping. There are attempts to derive the fraction of energy lost based on the change in amplitude.

Discussion Status

Some participants have provided calculations and expressed uncertainty about their correctness. There is an ongoing exploration of the mathematical relationships involved, with some guidance offered on simplifying the expressions.

Contextual Notes

Participants note a lack of specific information in textbooks regarding energy in damped oscillators, which leads to questions about the assumptions and formulas used in their calculations.

hemetite
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The amplitude of a lightly damped osciallator decreases by 3% during each cycle of oscillation. What fraction of the energy is lost in each cycle?

Okay..i couldn't find anything about energy in the damped oscillation is my textbook..

but in SHM sections. it showed that

Energy of the SHM system = (1/2)kAsquare

where A is the amplitude and k is the coefficient for the spring..

but for lightly damped osciallator...the Energy of the SHM will change for every cycle..in this case, it will drop by 3%...

i know that the period for every cycle is the same but only the amplitude will change.

here is my guess...

Energy of the SHM system = (1/2)kAsquare -----------> first cycle

Energy of the SHM system = (1/2)k(A*0.97)square -----------> second cycle

Energy lost = (1/2)kAsquare - (1/2)k(A*0.97)square ----------> difference

Fraction of energy lost = difference/(1/2)kAsquare

am i right here?...i am a feeling i am wrong..but dun know how to aprroach this..
 
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hemetite said:
Energy of the SHM system = (1/2)kAsquare -----------> first cycle

Energy of the SHM system = (1/2)k(A*0.97)square -----------> second cycle

Energy lost = (1/2)kAsquare - (1/2)k(A*0.97)square ----------> difference

Fraction of energy lost = difference/(1/2)kAsquare

am i right here?...i am a feeling i am wrong..but dun know how to aprroach this..

Hi hemetite! :smile:

(have a squared: ² :wink:)

I haven't checked whether your formula (1/2)kA² is right, but assuming it is …

yes, that's fine! (though a bit long-winded :wink:) … now divide everything by (1/2)kA², and the result is … ? :smile:
 
..well...i am new here...i worry if i never explained thouroughly...you guys won't help me...hahaha...

anyway...here goes

=[(1/2)kA² - (1/2)k(0.97A)²] / [(1/2)kA²]

= 0.0591

should be correct now...
 
hemetite said:
0.0591

should be correct now...

Sure is! :biggrin:
 
hahaha...man ..i love this forum..

people here are helpful and funny...
 
hemetite said:
people here are helpful and funny...

oi!
:biggrin: so are the fish! :biggrin:​
 

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