A pipe 20 cm long Is closed at one end. Which harmonic mode of the pipe 1s 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 ).
Length of the closed end pipe,
l = 20 cm = 0.2 m
Speed of sound, v = 340 ms-1
The fundamental frequency, v = v/4l = {340}/{4 x 0.2}
= 425 Hz
This is the first harmonic. The second harmonic is 3v i.e., 1275 Hz, the third harmonic is 5v i.e., 2125 Hz etc. Therefore only the first harmonic of frequency 425 Hz is resonantly excited by a 430 Hz source.
If both the ends of the pipe are open, the fundamental frequency is given by:
v’ = v/2l = {340}/{2 x 0.2} = 850 Hz
The second, third, fourth … etc. harmonics have frequencies 2v’, 3v’, 4v’, ….i.e., 1700 Hz, 2550 Hz, 3400 Hz. etc. No harmonic can be excited by the 430 Hz source in this case.
Given :
l = 20 cm = 0.2 m
v = 340 ms-1
f = 430 Hz
For a pipe closed at one end :
f = \(\frac{(2n-1)v}{4l}\) n = 1, 2, 3 ……
⇒ 430 = (2n-1)\(\frac{340}{4 \times 0.2}\)
⇒ n = 1
⇒ Resonance occurs only for first/fundamental mode of vibration. For a pipe open at both ends,
f = nv/2l n = 1, 2, …..
⇒ 430 = \(\frac{n \times 340}{2 \times 0.2}\)
⇒ n = 0.51
Since n<1, resonance does not occur.