Question

(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?

Short answer

a. The fundamental frequency is 256 Hz.

b. The second harmonic is 512 Hz.

Step-by-step solution

Given Data

The length of the tube is0.672 m.

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

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.

Calculation of the fundamental frequency 

(a)

The fundamental frequency is,

$f=\frac{v}{2l}$

Plugging the values,

$\begin{array}{r}f=\frac{344}{2\times 0.672}\\ f=256Hz\end{array}$

Therefore the fundamental frequency is 256 Hz.

Calculation of the second harmonic

(b)

The second harmonic is,

$f^{\prime }=2\times \frac{v}{2l}$

Plugging the values,

$\begin{array}{r}{f}^{\prime }=2\times \frac{344}{2\times 0.672}\\ f^{\prime }=512Hz\end{array}$

Therefore the frequency of its second harmonic is 512 Hz.