Правка разделов «Statement», «Solution», «Answer»
en/3.5.31.md
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| ### Statement | |||
| − | $3.5.31.$ [Insert the problem statement] | ||
| + | $3.5.31.$ When the forcing's frequency is small as compared to the oscillator's natural frequency, its displacement can be considered equal to $F(t)/k$, where $F(t)$ is the oscillating force, $k$ is the rigidity of the oscillating system. At high frequencies of the forcing force, the acceleration of the oscillator can be considered equal to $F(t)/m$, where $m$ is the mass of the oscillator. Explain this. | ||
| ### 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 $x\approx F(t)/m$. | ||
| + | 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 $a=\ddot x\approx F(t)/m$. | ||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | See the solution. | ||
| @@ -1,15 +1,15 @@ | |||
| ### Statement | ### Statement | ||
| $3.5.31.$ [Insert the problem statement] | $3.5.31.$ When the forcing's frequency is small as compared to the oscillator's natural frequency, its displacement can be considered equal to $F(t)/k$, where $F(t)$ is the oscillating force, $k$ is the rigidity of the oscillating system. At high frequencies of the forcing force, the acceleration of the oscillator can be considered equal to $F(t)/m$, where $m$ is the mass of the oscillator. Explain this. | ||
| ### Solution | ### 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 | 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 | ||
| \[m\ddot x+kx=F(t)=F_0\sin(\omega t),\] | \[m\ddot x+kx=F(t)=F_0\sin(\omega t),\] | ||
| 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 $x\approx F(t)/m$. | 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 $a=\ddot x\approx F(t)/m$. | ||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | See the solution. | ||