Новое решение
en/3.5.36.md
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| + | ### Statement | ||
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| + | $3.5.36.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | Substituting the particular (i.e. steady-state) solution $x=A\cos(\omega_0t-\phi)$ into the expression $\ddot x+2\gamma\dot x$+\omega_0^2x, we obtain $-2\gamma\omega A\sin(\omega_0t-\phi)$. (Note that the first and third terms cancel out, because the particular solution happens to be the solution to the homogeneous equation $\ddot x+\omega_0^2x=0$.) Thus, the required forcing is $F(t)=-2m\gamma\omega A\sin(\omega_0t-\phi)$. | ||
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| + | Conversely, if the forcing has amplitude $F_0$ and $\omega=\omega_0$, then $2m\gamma\omega A=F_0$, and the amplitude of the forced oscillation is $A=F_0/(2m\gamma\omega_0)$. When a constant force $F_0$ is applied instead, the deflection of the oscillator is $F_0/(m\omega_0^2)$. The amplitude of the forced oscillation is thus $\omega_0/(2\gamma)$ times this deflection. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.5.36.$ [Insert the problem statement] | |||
| ### Solution | |||
| Substituting the particular (i.e. steady-state) solution $x=A\cos(\omega_0t-\phi)$ into the expression $\ddot x+2\gamma\dot x$+\omega_0^2x, we obtain $-2\gamma\omega A\sin(\omega_0t-\phi)$. (Note that the first and third terms cancel out, because the particular solution happens to be the solution to the homogeneous equation $\ddot x+\omega_0^2x=0$.) Thus, the required forcing is $F(t)=-2m\gamma\omega A\sin(\omega_0t-\phi)$. | |||
| Conversely, if the forcing has amplitude $F_0$ and $\omega=\omega_0$, then $2m\gamma\omega A=F_0$, and the amplitude of the forced oscillation is $A=F_0/(2m\gamma\omega_0)$. When a constant force $F_0$ is applied instead, the deflection of the oscillator is $F_0/(m\omega_0^2)$. The amplitude of the forced oscillation is thus $\omega_0/(2\gamma)$ times this deflection. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||