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Q42PE

Expert-verifiedFound in: Page 630

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**(a) What is the fundamental frequency of a 0.672 m long tube, open at both ends, on a day when the speed of sound is 344 m/s? (b) What is the frequency of its second harmonic?**

a. The fundamental frequency is 256 Hz.

b. The second harmonic is 512 Hz**.**

The length of the tube is 0.672 m.

The speed of sound is v = 344 m/s.** **

**The expression for the fundamental frequency is given by,**

\(f = \frac{v}{{2l}}\)** **

**Here v is the velocity of the sound, l is the length of the wind instrument and f is frequency of sound.**

(a)

The fundamental frequency is,

\(f = \frac{v}{{2l}}\)

Plugging the values,

\(\begin{aligned}f = \frac{{344}}{{2 \times 0.672}}\\f = 256\;Hz\end{aligned}\)

Therefore the fundamental frequency is 256 Hz.

(b)

The second harmonic is,

\(f' = 2 \times \frac{v}{{2l}}\)

Plugging the values,

\(\begin{aligned}f' = 2 \times \frac{{344}}{{2 \times 0.672}}\\{f'} = 512\;Hz\end{aligned}\)

Therefore the frequency of its second harmonic is 512 Hz.

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