A) \[32.8\,\,cm\]
B) \[50\,\,cm\]
C) \[65.8\,\,cm\]
D) \[25\,\,cm\]
Correct Answer: A
Solution :
Given, \[f=256\,Hz\] \[{{v}_{sound}}=336\,m{{s}^{-1}}\] \[\lambda =\frac{v}{f}=\frac{336}{256}=1.3125m\] \[=131.25\,\,cm\] The distance between a node (N) and adjoining antinode \[=\frac{\lambda }{4}=\frac{131.25}{4}=32.8\,\,cm\] the shortest distance from the wall at which the air particles will have maximum amplitude of vibration is\[32.8\text{ }cm\].
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