Chapter 15, Problem 64A

To determine

To find: The length of the second pipe.

Answer to Problem 64A

The length of second pipe should be $6.70\times {10}^2\text{ mm}$ .

Explanation of Solution

Given data:

Length of the first pipe is $836\text{ mm}$ . The second one should have a pitch that is one major third higher.

Formula used:

We know that $\text{L=}\frac{\lambda }{2}$ , so $\lambda \text{= 2L and }\lambda \text{=}\frac{\text{v}}{f}$ .

Hence, the formula used here is $f=\frac{\text{v}}{2\text{L}}$

Calculation:

Substituting the values, we get

  $f=\frac{343\text{ m/s }}{(2)\left(0.836\text{ m}\right)}=\text{205Hz}$

The ratio of a frequency one major third higher is $\text{5:4}$ , so

$\text{(205 Hz)}\left(\frac{5}{4}\right)=256\text{Hz}$

The length of the second pipe is

  $\begin{array}{l}L\text{=}\frac{v}{2f}\\ =\frac{343\text{ m/s}}{\left(2\right)\left(256\text{ Hz}\right)}\\ =\text{ 6}\text{.70}\times {\text{10}}^2\text{mm}\end{array}$

Conclusion:

Length of the second pipe should be $6.70\times {10}^2\text{ mm}$ .