A tuba creates a 4th Harmonic of frequency 116.5 Hz. When the first valve is pushed, it opens an extra bit of tubing 0.721m long. What is the new frequency of the 4th Harmonic? Hz

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