Chapter 15.2, Problem 21SSC

(a)

To determine

The frequency of the note played by the pipe.

Answer to Problem 21SSC

The value of frequency of note played by the pipe is $35.73$ Hz.

Explanation of Solution

Given:

Length of the pipe, $L=2.4\text{ m}$

Speed of sound, $v=343\text{ m/s}$

Formula used:

The formula of frequency is,

  $f=\frac{v}{4L}$

Where, $v$ = speed of sound

  $L$ = length of pipe

Calculation:

  $\begin{array}{l}f=\frac{v}{4L}\\ f=\frac{343}{4\left(2.4\right)}\\ f=35.73\text{ Hz}\end{array}$

Conclusion:

The frequency of the note played by the pipe is $f=35.73\text{ Hz}$ .

(b)

To determine

The length of second pipe by which it is greater than the first pipe.

Answer to Problem 21SSC

The length of second pipe is greater than first pipe by $0.10\text{ m}$ .

Explanation of Solution

Given:

Speed of the sound, $v=343\text{ m/s}$

Frequency 1, $f_1=35.73\text{ Hz}$

Beat Frequency, $f_b=1.4\text{ Hz}$

Formula used:

  $\lambda =\frac{v}{f}$

  $L=\frac{\lambda }{4}$

Where, $v$ = speed of sound

  $\lambda$ = wavelength

Calculation:

  $\begin{array}{l}{f}_2=f_1-f_b\\ f_2=35.73-1.4\\ f_2=34.33\text{ Hz}\end{array}$

The wavelength is calculated as,

  $\begin{array}{l}\lambda =\frac{v}{f_2}\\ \lambda =\frac{343}{34.3}\\ \lambda =10\text{ m}\end{array}$

The length of second pipe is calculated as,

  $\begin{array}{l}{L}_2=\frac{\lambda }{4}\\ L_2=\frac{10}{4}\\ L_2=2.50\text{ m}\end{array}$

The difference in the length of second pipe and first pipe is,

  $\begin{array}{l}\Delta L=L_2-L_1\\ \Delta L=2.50-2.40\\ \Delta L=0.10\text{ m}\end{array}$

Conclusion:

The length of second pipe is greater than the first pipe by $0.10\text{ m}$ .