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