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 is maximum at antinode. The wall acts like a rigid boundary and reflects 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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