A pipe has a length of 1.29 m. Determine the frequency of the first harmonic if the pipe is open at each end. The velocity of sound in air is 343 m/s. Answer in units of Hz.

132.95 Hz.

Explanation:

Given:

v = 343 m/s

L = 1.29 m.

Since the pipe is open at both ends,

L = λ/2

λ = v/f = 2L

= 2 * 1.29

= 2.58 m

f = 343/2.58

= 132.95 Hz.

265.9Hz

Explanation:

In an open pipe, both ends of the pipes are opened. The fundamental frequency in an open pipe is expressed as fo = V/2L where;

f is the frequency of the wave

V is the velocity of the wave = 343m/s

L is the length of the pipe = 1.29m

Substituting the value to get the fundamental frequency in the open pipe we have;

Fo = 343/2 (1.29)

Fo = 343/2.58

Fo = 132.95Hz

Harmonics are integral multiples of the fundamental frequency e. g 2fo, 3fo, 4fo, 5fo ...

The first harmonic in the open pipe will be f1 = 2fo

Since f1 = 2 (132.95)

f1 = 265.9Hz

The frequency of the first harmonic if the pipe is open at each end is 265.9Hz