If I have a damped oscillator (with no driving force), the energy of the oscillator will decrease like: $$E(t)=E_0e^{-\gamma t},$$

where $E_0$ is some initial energy and $\gamma\in\mathbb R^ +$. We have that $\gamma=c\omega_0$, where $\omega_0=\sqrt{k/m}$ is the natural fequency of the oscillator.

I don't understand why the fraction of the energy lost per cycle is given by $E(T)/E_0=e^{-\gamma T}$, where $T=2\pi/\omega_0$ (This is what Vibrations and Waves by French says). I mean, if I ask the fraction of the energy lost per 10 cycles, this would be $e^{-\gamma10T}$. But this is greater than the energy lost in one cycle, so this doesn't make much sense to me. Could someone explain to me why the expression $E(T)/E_0=e^{-\gamma T}$ gives the fraction of energy lost per cycle?

Thanks a lot!


The period of the natural frequency is $T=2\pi/\omega_0$. Therefore the energy at the end of a cycle (measuring time from the start of the cycle) is:$E(T)=E(0)e^{-\gamma T}$. Hence the fraction of the energy lost per cycle is $F_1=E(T)/E(0)=e^{-\gamma T}$.

If the fraction of energy lost in 1 cycle is $F_1$ then in two cycles we lose a fraction $F_1$ on the first and $F_1$ again on the second. This gives a combined fractional loss of $F_2=F_1^2$. Repeating this for ten cycles gives the fractional loss over ten cycles of $F_{10}=F_1^{10}$ and as $F_1<1$ we have $F_{10}<F_1$.

Now if the fractionsl loss on one cycle is $F_1=e^{-\gamma T}$ we have: $F_{10}=F_1^{10}=\left(e^{-\gamma T}\right)^{10}=e^{-\gamma 10 T}$. So yes over ten cycles you do indeed lose more energy than over a single cycle.

  • $\begingroup$ I think I have a problem with the language ("fraction loss"), thanks anyway :D $\endgroup$ – Vladimir Vargas May 8 '15 at 2:08

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