Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

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:

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

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.