Question

A tuba creates a 4th harmonic of frequency 116.5 Hz. When the first valve is pushed, it opens an extra bit of tubing 0.721 m long. What is the new frequency of the 4th harmonic ? (Hint : Find the original length .)

Answer 1

Answer:

The new frequency is:

f₄ = 93.54 Hz

Explanation:

The data given to us is:

No. of harmonic frequency = 4

f₄ = 116.5 Hz

We know that the harmonic frequency is given by the formula:

fₙ = (n · v)/4L

where

n = no. of harmonic frequency

v = speed of sound

L = Length of rope

Lets find the initial length of rope for 4rth harmonic frequency (n=4) and v=343 m/s

fₙ = (n · v)/4L

f₄ = 4v/4L

f₄ = v/L

L = v/f₄

L = 343/116.5

L= 2.944m

Find frequency for new length that is 2.944m + 0.721m = 3.667m

f₄ = v/l

f₄ = 343/3.667

f₄ = 93.54 Hz