A) \[\text{2 }\!\!\times\!\!\text{ }\frac{\text{4}}{\text{330}}\text{Hz}\]
B) \[\text{2 }\!\!\times\!\!\text{ }\frac{330}{4}\text{Hz}\]
C) \[\text{3 }\!\!\times\!\!\text{ }\frac{330}{4}\text{Hz}\]
D) \[\text{4 }\!\!\times\!\!\text{ }\frac{330}{4}\text{Hz}\]
Correct Answer: C
Solution :
Length of closed organ pipe = 1 m Velocity of sound = 330 m/s As the frequencies of a closed organ pipe are \[\frac{\upsilon }{4l},\frac{3\upsilon }{4l},\frac{5\upsilon }{4l}\] Hence, the frequency for second note is \[=\frac{3\upsilon }{4l}=3\times \frac{330}{4}Hz\]
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