Answer to Question #269132 in Physics for sai

Question

A damped oscillator makes 100 oscillations per second. The energy decays to 1/e of

its initial value in 10 ms. How many oscillations per second will the oscillator make

in the absence of damping.

Answer 1

$\nu=\frac{\sqrt{\omega_0^2-\beta^2}}{2\pi}\to 4\pi ^2\nu^2=\omega_0^2-\beta^2\to4\pi ^2\nu^2+\beta^2=\omega_0^2$

$\omega_0=\sqrt{4\pi ^2\nu^2+\beta^2}$

$\nu_0=\frac{\omega_0}{2\pi}=\frac{\sqrt{4\pi ^2\nu^2+\beta^2}}{2\pi}$

$E_2/E_1=A_2^2/A_1^2\to (E_1/e)/E_1=A_2^2/A_1^2\to 1/e=A_2^2/A_1^2$

$e=A_1^2/A_2^2=(A_0e^{-\beta t_0}/A_0e^{-\beta t})^2=(1/e^{-\beta t})^2$ ; $t_0=0$

$e=e^{2\beta t}\to \beta=1/(2t)=1/(2\cdot0.01)=50\ (s^{-1})$

$\nu_0=\frac{\sqrt{4\pi ^2\nu^2+\beta^2}}{2\pi}=\frac{\sqrt{4\cdot3.14^2\cdot 100^2+50^2}}{2\cdot3.14}=100.32\ (Hz)$