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en/3.5.31.md
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| + | ### Statement | ||
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| + | $3.5.31.$ [Insert the problem statement] | ||
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| + | ### Solution | ||
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| + | Assuming that the forced oscillation dominates the free oscillation, when the particular solution $x=A\sin(\omega t)$ is substituted into the equation of motion | ||
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| + | \[m\ddot x+kx=F(t)=F_0\sin(\omega t),\] | ||
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| + | we see that the relative magnitude of the two terms on the left are $m\omega^2:k=\omega^2:\omega_0^2$. If $\omega\ll\omega_0$, then $kx\approx F(t)$, so $x\approx F(t)/k$. On the other hand, if $\omega\gg\omega_0$, then $m\ddot x\approx F(t)$, so $x\approx F(t)/m$. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.5.31.$ [Insert the problem statement] | |||
| ### Solution | |||
| Assuming that the forced oscillation dominates the free oscillation, when the particular solution $x=A\sin(\omega t)$ is substituted into the equation of motion | |||
| \[m\ddot x+kx=F(t)=F_0\sin(\omega t),\] | |||
| we see that the relative magnitude of the two terms on the left are $m\omega^2:k=\omega^2:\omega_0^2$. If $\omega\ll\omega_0$, then $kx\approx F(t)$, so $x\approx F(t)/k$. On the other hand, if $\omega\gg\omega_0$, then $m\ddot x\approx F(t)$, so $x\approx F(t)/m$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||