Question #269132

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.


1
Expert's answer
2021-11-21T17:27:53-0500

ν=ω02β22π4π2ν2=ω02β24π2ν2+β2=ω02\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


ω0=4π2ν2+β2\omega_0=\sqrt{4\pi ^2\nu^2+\beta^2}


ν0=ω02π=4π2ν2+β22π\nu_0=\frac{\omega_0}{2\pi}=\frac{\sqrt{4\pi ^2\nu^2+\beta^2}}{2\pi}



E2/E1=A22/A12(E1/e)/E1=A22/A121/e=A22/A12E_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=A12/A22=(A0eβt0/A0eβt)2=(1/eβt)2e=A_1^2/A_2^2=(A_0e^{-\beta t_0}/A_0e^{-\beta t})^2=(1/e^{-\beta t})^2 ; t0=0t_0=0


e=e2βtβ=1/(2t)=1/(20.01)=50 (s1)e=e^{2\beta t}\to \beta=1/(2t)=1/(2\cdot0.01)=50\ (s^{-1})


ν0=4π2ν2+β22π=43.1421002+50223.14=100.32 (Hz)\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)









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