Chapter 15, Problem 56A

To determine

To calculate:The frequency of the tuning fork.

Answer to Problem 56A

The frequency of the tuning fork is $536\text{ Hz}$ .

Explanation of Solution

Given info:

A tuning fork vibrates over the mouth of a vertical tube filled with water.

First resonance length ( $L_1$ ) = $17\text{ cm}$

Second resonance length ( $L_2$ ) = $\text{49 cm}$

Formula used:

The distance between water levels is equal to the distance between two successive nodes of half of a wavelength.

  $\frac{\lambda }{2}=L_2-L_1$

$\lambda$ represents the wavelength of the resonant sound, $L_1$ represents the first resonance length and $L_2$ represents the second resonance length.

In addition, the frequency is given by

  $f=\frac{v}{\lambda }$

$f$ represents the frequency of the tuning fork, $v$ represents the speed of sound in air and $\lambda$ represents the wavelength of the resonant sound.

Calculation:

As per the problem

First resonance length ( $L_1$ ) = $17\text{ cm}$

Second resonance length ( $L_2$ ) = $\text{49 cm}$

Wavelength of the resonant sound

  $\frac{\lambda }{2}=L_2-L_1$

  $\begin{array}{l}\frac{\lambda }{2}=49-17\\ \lambda =2(49-17)\\ \lambda =64\text{ cm}\\ \lambda \text{= 0}\text{.64 m}\end{array}$

Calculate frequency using the wavelength calculated above and take speed of sound in air $v=343\text{ m/s}$

  $\begin{array}{l}f=\frac{v}{\lambda }\\ f=\frac{343\text{ m/s}}{0.64\text{ m}}=536\text{ Hz}\end{array}$

Conclusion:

Thus, thefrequency of the tuning fork is $536\text{ Hz}$ .