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College Physics (Urone)
Found in: Page 630
College Physics (Urone)

College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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Short Answer

(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.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Given Data

The length of the tube is 0.672 m.

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

Step 2: Calculation of the fundamental frequency

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.

Step 3: Calculation of the fundamental frequency 

(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.

Step 4: Calculation of the second harmonic

(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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