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pochemuha
3 years ago
14

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.
Physics
1 answer:
yulyashka [42]3 years ago
4 0

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

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Answer:

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<h3>What is the maximum diffraction order seen?</h3>

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