A tube that is open on both ends is 4 meters long. Assuming that the speed of sound is 340m/s, find the following:
a) What is the wavelength of the fundamental standing wave (the first harmonic)
b) What is the frequency of the fundamental wave?
c) What is the frequency of the second harmonic?
See previous post: Mon., 5-8-17, 2:43 PM.
To solve this problem, we need to use the formula for the wavelength of a sound wave in a tube that is open at both ends:
λ = 2L/n
where λ is the wavelength, L is the length of the tube, and n is the harmonic number.
a) To find the wavelength of the fundamental standing wave (first harmonic), we substitute L = 4m and n = 1 into the formula:
λ = 2(4m) / 1
λ = 8m
So, the wavelength of the first harmonic is 8 meters.
b) To find the frequency of the fundamental wave, we need to use the formula for the speed of sound:
v = fλ
where v is the speed of sound and f is the frequency.
We can rearrange this formula to solve for the frequency:
f = v/λ
Substituting the values v = 340m/s and λ = 8m:
f = 340m/s / 8m
f = 42.5 Hz
So, the frequency of the fundamental wave is 42.5 Hz.
c) To find the frequency of the second harmonic, we use the formula:
f2 = f1 * 2
where f2 is the frequency of the second harmonic and f1 is the frequency of the first harmonic.
So, substituting the known value of f1 = 42.5 Hz:
f2 = 42.5 Hz * 2
f2 = 85 Hz
Therefore, the frequency of the second harmonic is 85 Hz.