A) \[\frac{7}{8}m\]
B) \[\frac{3}{8}m\]
C) \[\frac{1}{8}m\]
D) \[\frac{1}{4}m\]
Correct Answer: C
Solution :
Key Idea: Amplitude 15 maximum at antinode. The wall acts like a rigid boundary and reflect this wave and sends it back towards the open end. At the open end an antinode is formed and a node is formed at the wall. The distance between antinode and node is\[\frac{\lambda }{4}.\] Therefore, if n be the frequency of note emitted then \[\lambda =\frac{v}{n}\] \[\Rightarrow \] \[\lambda =\frac{300}{600}=\frac{1}{2}m\] Maximum amplitude is obtained at distance \[=\frac{\lambda }{4}=\frac{1}{2}\times \frac{1}{4}=\frac{1}{8}m\]
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