A) 1
B) \[\frac{5}{6}\]
C) \[\frac{3}{4}\]
D) \[\frac{6}{7}\]
Correct Answer: D
Solution :
At \[{{x}_{1}}=\frac{\pi }{3k}\] and \[{{x}_{2}}=\frac{3\pi }{2k}\] |
\[{{\operatorname{sinkx}}_{1}}\] or \[{{\operatorname{sinkx}}_{2}}\] is not zero. |
Therefore, neither of \[{{x}_{1}}\] or \[{{x}_{2}}\]is a node |
\[\Delta x={{x}_{2}}-{{x}_{1}}=\left( \frac{3}{2}-\frac{1}{3} \right)\frac{\pi }{k}=\frac{7\pi }{6k}\] |
Since \[\frac{2\pi }{k}>\Delta x>\frac{\pi }{k};\,\lambda >\Delta x>\frac{\lambda }{2}\] |
\[\left( k=\frac{2\pi }{\lambda } \right)\] |
Therefore, \[{{\phi }_{1}}=\pi \,and\,{{\phi }_{2}}=k.\Delta x=\frac{7\pi }{6};\therefore \frac{{{\phi }_{1}}}{{{\phi }_{2}}}=\frac{6}{7}\] |
NOTE: In case of a stationary wave phase difference between any two points is either zero or \[\pi \]. |
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