Правка разделов «Statement», «Solution», «Answer»
en/3.9.9.md
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| @@ -1,6 +1,6 @@ | |||
| ### Statement | |||
| − | $3.9.9.$ [Insert the problem statement] | ||
| + | $3.9.9.$ Two external longitudinal forces act in two sections of an infinite rod. The force on the left side changes according to the law $F_1=F_0\cos\omega t$, and the force on the right side changes according to the law $F_2=F_0\sin\omega t$. At what distance $l$ between the sources of the force will the traveling wave propagate from left to right only or from right to left only? | ||
| ### Solution | |||
| Let $x=0$ and $x=l$ be the positions of the left and the right sources, respectively. Then, the superposition of the waves generated by both sources is given by | |||
| \[A[\cos(\omega t-k|x|)+\sin(\omega t-k|x-l|)],\] | |||
| @@ -10,16 +10,18 @@Solution | |||
| where $k=\omega/c$ is the wave number and $c$ is the speed of the wave. Clearly, a standing wave is formed for $0\le x\le l$. In order that left traveling wave for $x<0$ vanishes and only right traveling wave for $x>l$ remains, we must have | |||
| − | \[\cos(\omega t+kx)+\sin(\omega t+k(x-l))=0\qquad\Rightarrow\qquad kl=n | ||
| + | \[\cos(\omega t+kx)+\sin(\omega t+k(x-l))=0\qquad\Rightarrow\qquad kl=n\cdot2\pi+\frac{\pi}{2},\] | ||
| − | where $n$ is an integer. | ||
| + | where $n$ is an integer. Thus, $l=(n+1/4)\lambda$, where $\lambda=2\pi c/\omega$. | ||
| On the other hand, in order that right traveling wave for $x>l$ vanishes and only left traveling wave for $x<0$ remains, we must have | |||
| − | \[\cos(\omega t-kx)+\sin(\omega t-k(x-l))=0\qquad\Rightarrow\qquad kl=n | ||
| + | \[\cos(\omega t-kx)+\sin(\omega t-k(x-l))=0\qquad\Rightarrow\qquad kl=n\cdot2\pi+\frac{3\pi}{2}.\] | ||
| Thus, $l=(n+3/4)\lambda$. | |||
| #### Answer | |||
| − | |||
| + | Only right traveling wave remains when $l=(n+1/4)\lambda$, where $\lambda=2\pi c/\omega$. | ||
| + | |||
| + | Only left traveling wave remains when $l=(n+3/4)\lambda$. | ||
| @@ -1,6 +1,6 @@ | |||
| ### Statement | ### Statement | ||
| $3.9.9.$ [Insert the problem statement] | $3.9.9.$ Two external longitudinal forces act in two sections of an infinite rod. The force on the left side changes according to the law $F_1=F_0\cos\omega t$, and the force on the right side changes according to the law $F_2=F_0\sin\omega t$. At what distance $l$ between the sources of the force will the traveling wave propagate from left to right only or from right to left only? | ||
| ### Solution | ### Solution | ||
| Let $x=0$ and $x=l$ be the positions of the left and the right sources, respectively. Then, the superposition of the waves generated by both sources is given by | Let $x=0$ and $x=l$ be the positions of the left and the right sources, respectively. Then, the superposition of the waves generated by both sources is given by | ||
| \[A[\cos(\omega t-k|x|)+\sin(\omega t-k|x-l|)],\] | \[A[\cos(\omega t-k|x|)+\sin(\omega t-k|x-l|)],\] | ||
| @@ -10,16 +10,18 @@Solution | |||
| where $k=\omega/c$ is the wave number and $c$ is the speed of the wave. Clearly, a standing wave is formed for $0\le x\le l$. In order that left traveling wave for $x<0$ vanishes and only right traveling wave for $x>l$ remains, we must have | where $k=\omega/c$ is the wave number and $c$ is the speed of the wave. Clearly, a standing wave is formed for $0\le x\le l$. In order that left traveling wave for $x<0$ vanishes and only right traveling wave for $x>l$ remains, we must have | ||
| \[\cos(\omega t+kx)+\sin(\omega t+k(x-l))=0\qquad\Rightarrow\qquad kl=n |
\[\cos(\omega t+kx)+\sin(\omega t+k(x-l))=0\qquad\Rightarrow\qquad kl=n\cdot2\pi+\frac{\pi}{2},\] | ||
| where $n$ is an integer. |
where $n$ is an integer. Thus, $l=(n+1/4)\lambda$, where $\lambda=2\pi c/\omega$. | ||
| On the other hand, in order that right traveling wave for $x>l$ vanishes and only left traveling wave for $x<0$ remains, we must have | On the other hand, in order that right traveling wave for $x>l$ vanishes and only left traveling wave for $x<0$ remains, we must have | ||
| \[\cos(\omega t-kx)+\sin(\omega t-k(x-l))=0\qquad\Rightarrow\qquad kl=n |
\[\cos(\omega t-kx)+\sin(\omega t-k(x-l))=0\qquad\Rightarrow\qquad kl=n\cdot2\pi+\frac{3\pi}{2}.\] | ||
| Thus, $l=(n+3/4)\lambda$. | Thus, $l=(n+3/4)\lambda$. | ||
| #### Answer | #### Answer | ||
| Only right traveling wave remains when $l=(n+1/4)\lambda$, where $\lambda=2\pi c/\omega$. | |||
| Only left traveling wave remains when $l=(n+3/4)\lambda$. | |||