Answer to Question #89778 in Physics for Shivam Nishad

Question

A pipe of length 20 cm is closed at one end.
Which harmonic mode of the pipe is in
resonance with a source of frequency
425 Hz ? Speed of sound = 340 ms^-1.

Answer 1

Solution:

For a tube with one open end and one closed end all frequencies

$$f_n=nv/(4L) = nf_1$$

with n equal to an odd integer are natural frequencies, i.e. only odd harmonics of the fundamental are natural frequencies.

In our case:

$$L= 20 cm = 0.2m$$ $$f_n=425 Hz$$ $$v=340 m/s$$

So,

$$n=4Lf_n/v$$ $$n = (4 * 0.2*425)/340 = 1$$

Answer: 1 , first harmonic mode is in resonance with given frequency

Reference: https://en.wikipedia.org/wiki/Acoustic_resonance#Closed_at_one_end