Wave Motion 2 Question 38
38. Sound waves of frequency $660 Hz$ fall normally on a perfectly reflecting wall. The shortest distance from the wall at which the air particles have maximum amplitude of vibration is ….. m. Speed of sound $=330 m / s$.
(1984, 2M)
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Answer:
Correct Answer: 38. 0.125
Solution:
- Wall will be a node (displacement). Therefore, shortest distance from the wall at which air particles have maximum amplitude of vibration (displacement antinode) should be $\lambda / 4$. Here, $\lambda=\frac{v}{f}=\frac{330}{660}=0.5 m$
$\therefore$ Desired distance is $\frac{0.5}{4}=0.125 m$
