Damped oscilator - logarithmic decrement of damping [closed]

Question

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?

Answer 1

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:

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