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Found in: Page 630

### College Physics (Urone)

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.

See the step by step solution

## 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.