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iogann1982 [59]
3 years ago
9

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
Physics
1 answer:
Ksenya-84 [330]3 years ago
6 0

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

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