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A pipe closed at one end and open at the other end resonates with a sound of frequency 135 Hz and also with 165 Hz, but not at any other frequency intermediate between these two. Then, the frequency of the fundamental note of the pipe isChoose the correct answer:15 Hz60 Hz7.5 Hz30 Hz

Chirag , 7 Years ago
Grade 12
anser 2 Answers
Saurabh Koranglekar

Last Activity: 4 Years ago

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Vikas TU

Last Activity: 4 Years ago

We know , frequency of sound wave in pipe closed at one end and open at the other end resonates, .
Here, v = velocity of speed.
         L = length of tube.
Now, it is given that their is no any wave between 135 Hz and 165 Hz.
Therefore, these are consecutive resonating frequencies.
So, 135 = (2n+1)v/4L
165 = (2(n+1)+1)v/4L = (2n+3)v/4L

Subtracting them we get ,
v/4L = 15
Therefore, fundamental frequency is 15.
Hence, this is the required solution.
 

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