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Question

A pipe of length 85cm is closed from one end. Find the number of possible natural oscillations of the air column in the pipe whose frequencies lie below 1250Hz. The velocity of sound in the air is 340ms.


A

6

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B

4

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C

12

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D

8

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Solution

The correct option is A

6


Step 1. Given Data,

Length of pipe, L=85cm=0.85m

Velocity of sound in air, v=340ms

Step 2. Fundamental frequency of closed organ pipe is,

f=v4L (where, f is the fundamental frequency.)

f=340(4×0.85)f=100Hz.

Step 3. To calculate the frequency:

f=vλf=340(4×0.85)f=100Hz.

The natural frequencies of closed Organ pipe is, fn=2n-1f

That is for n=1,2,3,4,5,6,7,

The value of fn=100Hz,300Hz,500Hz,700Hz,900Hz,1100Hz,1300Hz,

Then the possible frequency is100Hz,300Hz,500Hz,700Hz,900Hz,1100Hz below 1250Hz.

Thus the number of possible natural oscillations whose frequency is below 1250 is 6.

Hence, the correct option is (A).


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