Question

A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 88.4 Hz, determine the first three overtones. Use 343 m/s as the speed of sound in air.

Answer 1

Answer with Explanation:

We are given that

Fundamental frequency,f=88.4 Hz

Speed of sound,v=343 m/s

We have to find the first three overtones.

Tube is closed

The overtone of closed pipe  is equal to odd number of fundamental frequency.

Therefore, the overtone of tuba

$f'=nf$

Where n=3,5,7,..

Substitute n=3

$f'=3\times 88.4=265.2Hz$

For second overtone

$f'=5\times 88.4=442Hz$

For third overtone

$f'=7\times 88.4=618.8Hz$