Answer: New frequency = 103.3Hz
Explanation:
Given that the
Frequency F = 116.5 Hz
The tube is an opened tube.
The relationship between wavelength and the length of the tube is
λ = 2L/n
Where
λ = wavelength
L = length of the tube
n = number of harmonic
For 4th harmonic,
λ = 2L/4 = L/2
But wave speed V = F(λ)
Substitutes (λ) into the wave speed formula
V = FL/2
Let's assume that the V is the speed of sound = 330m/s
Substitutes the V and F into the formula above.
330 = 116.5L/2
Cross multiply
660 = 116.5L
L = 660/116.5
L = 5.67 m
Given that the first valve is pushed, it opens an extra bit of tubing 0.721m long.
New length = 5.67 + 0.721 = 6.39 m
New frequency will be
330 = 6.39F/2
Cross multiply
660 = 6.39F
F = 660/6.39
F = 103.3 Hz
