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Could you please tell me, where is the mistake?

What is the logarithmic decrement of damping $Λ$ of damped harmonic oscillator, if its mechanical energy decreases to the 50% of its initial value during first 10 seconds? The period of oscillations is T=2s. [Result: 0.0693].

Formula from wikipedia: $$ Λ = \frac{1}{n} \cdot ln\left( \frac{x(t)}{x(t + nT)} \right) $$

My solution:
Since I dont have amplitude and mass of particle, I have to work with ratio $ \frac{1}{\frac{1}{2}} $.

Since T=2s and elapsed time is 10seconds, $ n = 5 $. That gives me:

$$ Λ = \frac{1}{\frac{t}{T}} \cdot ln\left( \frac{1}{\frac{1}{2}} \right) $$ $$ Λ = \frac{1}{5} \cdot ln\left( \frac{1}{\frac{1}{2}} \right) $$ $$ Λ = 0.1386 $$

Which is twice as much as was expected as result. So where is a mistake?
It's clear that it should be:

$$ Λ = \frac{1}{10} \cdot ln\left( \frac{1}{\frac{1}{2}} \right) $$

but why?

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Energy is amplitude squared, and squaring multiplies the log by 2. –  John Rennie Apr 30 '13 at 6:27
    
i was wondering what a dumped oscillator was... –  nervxxx Apr 30 '13 at 7:15
    
@JohnRennie I have tried that, and it doesn not give the right result. Result is: 0.277. Should be: 0.0693. –  Fidilip Apr 30 '13 at 7:25
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1 Answer

up vote 0 down vote accepted

As John Rennie already told mechanical energy is proportional to the amplitude squared so if the ratios of energies is ${1 \over 2}$, the ratio of amplitudes will be ${1 \over \sqrt 2}$ then:

$\delta = {1 \over 5} ln{1 \over {1 \over \sqrt 2}} = {1 \over 5} ln\sqrt 2 = 0.0693$

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Aha! So not tu square it, but square-root it :). Thnak you for your answer. –  Fidilip Apr 30 '13 at 8:04
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