Question

A trumpet plays its 3rd harmonic at 510 Hz. It then opens a valve, which adds 0.110 m to its length. What is the new 3rd harmonic frequency? (Hint: Find the original length.) (Speed of sound = 343 m/s) (Unit = Hz)​

Answer 1

f = 459.8 Hz

Step-by-step explanation:

When a trumpet is playing it is an open tube at both ends, therefore there is a belly in each one, the resonance occurs to

λ = 2L 1st harmonic

λ = 2L /2 2nd harmonic

λ = 2L /3 3rd harmonic

the speed of the wave is

v = λ f

λ = v / f

we substitute in the third harmonic

`(v)/(f) = (2L)/(3)` (1)

L =
`(3)/(2) \ (v)/(f)`

L =
`(3)/(2) \ (343)/( 510)`

L = 1.009 m

indicates that to add ΔL = 0.110 m, so the total length is

L_total = L + ΔL

L _total = 1.009 + 0.110

L _total = 1,119 m

we use equation 1

f =
`(3)/(2) \ (v)/(L_(total))`

f =
`(3)/(2) \ (343)/(1.119)`

f = 459.8 Hz