Question

A pipe 20 cm long is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source ? Will the same source be in resonance with the pipe if both ends are open? (speed of sound in air is 340 m s¯¹).

Answer 1

Given, the length of the pipe is 20 cm, the frequency of resonance is 430 Hz and the speed of sound in air is 340 m/s .

The formula to calculate the frequency in a closed end pipe is,

f= ( 2n−1 )v 4l ( 2n−1 )= 4fl v

Here, the frequency of resonance is f, the length of the pipe is l and the speed of sound is v.

Substituting the values in the above equation, we get:

( 2n−1 )= 4( 430 )( 20× 1 m 10 3  cm ) 340 ( 2n−1 )=1.02 2n=2.02 n=1.01

Thus, the resonance will occur only for the first node of vibration.

The formula to calculate frequency in an open end pipe is,

f= nv 2l n= 2fl v

Here, the frequency of resonance is f, the length of the pipe is l and the speed of sound is v.

Substituting the values in the above equation, we get:

n= 2( 430 )( 20× 1 m 10 3  cm ) 340 =0.51

As the value of n obtained from the above calculations is less than one, so the resonance position cannot be obtained.

Thus, the resonance position cannot be obtained for an open pipe.