Question

5. 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? Why? [Speed of sound in the air
= 340 m s-1]
​

Answer 1

Length of the pipe, 1=20cm=0.2m

Source frequency =nth normal mode of frequency, fn =430Hz

Speed of sound, v = 340 m/s

In a closed pipe, the nth normal mode of frequency is given by the relation:

$f = (2n - 1) \frac{v}{4l} \: \: \:        n∈{1,2,3,....} \\ 430=(2n−1) \frac{340}{4 \times 0.2} \\ </p><p>2n−1= \frac{430 \times4\times0.2 }{340} = 1.01 \\ 2n = 2.01 \\ n =1$

Hence, the first mode of vibration frequency is resonantly excited by the given source.

In a pipe open at both ends, the nth mode of vibration frequency is given by the relation: 

fn=nv/2l

n=2lfn/v≈0.5

The same source will not be in resonance with the same pipe open at both ends.