Новое решение
en/3.5.30.md
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| + | ### Statement | ||
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| + | $3.5.30.$ [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 $\omega<\omega_0$, then $kx$ has the same sign as $F(t)$; that is, the displacement and the forcing has the same direction. On the other hand, if $\omega>\omega_0$, then $m\ddot x$ has the same sign as $F(t)$, so $kx$ and $F(t)$ have opposite signs; that is, the displacement and the forcing has the opposite directions. | ||
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| + | #### Answer | ||
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| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.5.30.$ [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 $\omega<\omega_0$, then $kx$ has the same sign as $F(t)$; that is, the displacement and the forcing has the same direction. On the other hand, if $\omega>\omega_0$, then $m\ddot x$ has the same sign as $F(t)$, so $kx$ and $F(t)$ have opposite signs; that is, the displacement and the forcing has the opposite directions. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||