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en/3.9.9.md
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| + | ### Statement | ||
| + | |||
| + | $3.9.9.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
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| + | 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 | ||
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| + | \[A[\cos(\omega t-k|x|)+\sin(\omega t-k|x-l|)],\] | ||
| + | |||
| + | 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 | ||
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| + | \[\cos(\omega t+kx)+\sin(\omega t+k(x-l))=0\qquad\Rightarrow\qquad kl=n*2\pi+\frac{\pi}{2},\] | ||
| + | |||
| + | where $n$ is an integer. Consequently, $l=(n+1/4)\lambda$, 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 | ||
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| + | \[\cos(\omega t-kx)+\sin(\omega t-k(x-l))=0\qquad\Rightarrow\qquad kl=n*2\pi+\frac{3\pi}{2}.\] | ||
| + | |||
| + | Thus, $l=(n+3/4)\lambda$. | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $3.9.9.$ [Insert the problem statement] | |||
| ### 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 | |||
| \[A[\cos(\omega t-k|x|)+\sin(\omega t-k|x-l|)],\] | |||
| 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 | |||
| \[\cos(\omega t+kx)+\sin(\omega t+k(x-l))=0\qquad\Rightarrow\qquad kl=n*2\pi+\frac{\pi}{2},\] | |||
| where $n$ is an integer. Consequently, $l=(n+1/4)\lambda$, 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 | |||
| \[\cos(\omega t-kx)+\sin(\omega t-k(x-l))=0\qquad\Rightarrow\qquad kl=n*2\pi+\frac{3\pi}{2}.\] | |||
| Thus, $l=(n+3/4)\lambda$. | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||