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

Concept :- In closed organ pipe only odd nber of harmonics are produced , whereas in case of open organ pipe is all even as well as odd harmonics are produced. Also fundamental frequency is the minimum frequency of vibration.

Here,
Length of the pipe ( L) = 20 cm
= 20 × 10^-2 = 0.2 m
Frequency of the source ( f) = 430 Hz
speed of sound = 340 m/s

Fundamental frequency in closed organ pipe ( fo) = V/4L
= 340/4 × 0.2 = 425 Hz
Hence, the fundamental frequency of closed organ pipe will resonate with the source.

In open pipe ,
Fundamental frequency = V/2L
= 340/2 × 0.2
= 850 Hz
Hence, There will be no resonance .

Answer 2

Solution :

Here, L=20cm=0.2m,vn=430Hz,υ=340ms−1

The frequency of nth normal mode of vibration of closed pipe is

vn=(2n−1)υ4L∴430=(2n−1)3404×0.2

2n−1=430×4×0.2340=1.02

2n=2.02,n=1.01

Hence it will be the 1st normal mode of vibration.

In a pipe, open at both ends, we have vn=n×υ2L=n×3402×0.2=430∴n=430×2×0.2340=0.5

As n has to be an integer, therefore, open organ pipe cnnot be in resonance with the source .