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# Answer to Question #103276 in Mechanics | Relativity for sakshi

Question #103276
Show that the average energy of a weakly damped oscillator is given by:
< E > = E0 exp (-2bt)
1
2020-02-27T09:58:23-0500

The energy of the system is

"E(t)=KE(t)+U(t)=\\frac{1}{2}m\\bigg(\\frac{dx}{dt}\\bigg)^2+\\frac{1}{2}kx^2."

The instantaneous displacement of a weakly damped harmonic oscillator is

"x (t) = a_0 \\text{exp} (- bt)\\cdot \\text{cos} (\\omega_dt + \\phi),\\\\\n\\space\\\\\n\\frac{dx(t)}{dt}= - a_0 \\text{exp} (- bt) [b \\text{cos} (\\omega_dt + \\phi) + \\omega_d\\text{sin} (\\omega_dt + \\phi)],\\\\\n\\space\\\\\nKE(t)=\\frac{1}{2}ma_0^2 \\text{exp}(-2bt) [b\\text{cos} (\\omega_dt + \\phi) + \\omega_d\\text{sin} (\\omega_dt + \\phi)]^2"

(open the squared parentheses).

The potential energy is

"U=\\frac{1}{2}kx^2=\\\\\n\\space\\\\\n=\\frac{1}{2}ka_0^2 \\text{exp} (-2bt)\\cdot \\text{cos}^2 (\\omega_dt + \\phi)."

Remember that

"k=m\\omega_0^2."

"U=\\frac{1}{2}kx^2=\\\\\n\\space\\\\\n=\\frac{1}{2}m\\omega_0^2a_0^2 \\text{exp} (-2bt)\\cdot \\text{cos}^2 (\\omega_dt + \\phi)."

During one oscillation, for small amplitudes, the average values of sine and cosine squared are equal

"\\text{sin}^2 (\\omega_dt + \\phi)=\\text{cos}^2 (\\omega_dt + \\phi)=\\frac{1}{2}."

Remember that the average values of sine per one oscillation is 0.

Therefore, after adding the KE with opened parentheses and U, we get

"<E>=\\frac{1}{2}ma_0^2 \\text{exp}(-2bt)\\bigg[\\frac{b^2+\\omega_0^2}{2}+\\frac{\\omega_d^2}{2}\\bigg]=\\\\\n\\space\\\\\n=\\frac{1}{2}ma_0^2\\omega^2_0\\text{exp}(-2bt)."

Since

"\\frac{1}{2}ma_0^2\\omega^2_0=E_0,"

we can finally write

"<E>=E_0\\text{exp}(-2bt)."

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