# Damped oscilator - logarithmic decrement of damping [closed]

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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## closed as off-topic by ACuriousMind, Kyle Kanos, JamalS, David Z♦Apr 5 '15 at 20:36

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## 1 Answer

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