A pipe open at both ends has a length of 0.8 meters. The speed of sound in air is 340 m/s340m/s.
1.
Calculate the fundamental frequency (first harmonic) of the pipe.
2.
Determine the frequency of the third harmonic.
3.
What is the wavelength of the third harmonic?
Solution:
Step 1: Calculate the Fundamental Frequency
For a pipe open at both ends, the fundamental frequency (first harmonic) has a wavelength λ1λ1 such that the length of the pipe LL is half the wavelength:
L=λ12L=2λ1
Given L=0.8 mL=0.8m:
0.8=λ120.8=2λ1 λ1=2×0.8=1.6 mλ1=2×0.8=1.6m
The fundamental frequency f1f1 is given by:
f1=vλ1f1=λ1v
Where vv is the speed of sound in air (340 m/s340m/s):
f1=3401.6f1=1.6340 f1=212.5 Hzf1=212.5Hz
Step 2: Determine the Frequency of the Third Harmonic
For a pipe open at both ends, the nth harmonic has a wavelength λnλn such that:
λn=2Lnλn=n2L
For the third harmonic (n=3n=3):
λ3=2×0.83λ3=32×0.8 λ3=1.63λ3=31.6 λ3≈0.533 mλ3≈0.533m
The frequency of the third harmonic f3f3 is given by:
f3=vλ3f3=λ3v
Given v=340 m/sv=340m/s and λ3≈0.533 mλ3≈0.533m:
f3=3400.533f3=0.533340 f3≈637.5 Hzf3≈637.5Hz
Step 3: Calculate the Wavelength of the Third Harmonic
