Edits to “Statement”, “Answer”
en/3.5.30.md
+2 −2
| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $3.5.30.$ [Insert the problem statement] | ||
| + | $3.5.30.$ Why do the directions of displacement and the driving force coincide when the oscillator is forced to oscillate at a frequency smaller than its natural frequency, and are opposite when its frequency is higher than its natural frequency? | ||
| ### 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. | |||
| @@ -12,4 +12,4 @@Solution | |||
| #### Answer | |||
| − | [Insert a concise answer or boxed result] | ||
| + | See the solution. | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $3.5.30.$ [Insert the problem statement] | $3.5.30.$ Why do the directions of displacement and the driving force coincide when the oscillator is forced to oscillate at a frequency smaller than its natural frequency, and are opposite when its frequency is higher than its natural frequency? | ||
| ### 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 $\omega<\omega_0$, 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. | 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. | ||
| @@ -12,4 +12,4 @@Solution | |||
| #### Answer | #### Answer | ||
| [Insert a concise answer or boxed result] | See the solution. | ||