Answer:
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
Answer:
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.
