Chapter 15.2, Problem 14PP

To determine

To calculate:The frequency of the tuning fork.

Answer to Problem 14PP

The frequency of the tuning fork is $859\text{ Hz}$ .

Explanation of Solution

Given info:

Speed of sound in air at ${27}^{\text{o}}\text{C}$ , $v=347\text{ m/s}$

Resonance spacing = $\text{20}\text{.2 cm}$

Formula used:

The resonance lengths in a closed pipe are found at half-wavelength intervals.

Hence, the length-wavelength relationship:

  $\frac{\lambda }{2}=L_2-L_1$

  $\lambda$ represents the wavelength of the resonant sound, $L_2-L_1$ represents the resonance spacing.

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 problem,

Speed of sound in air at ${27}^{\text{o}}\text{C}$ ( $v$ ) = $347\text{ m/s}$

Resonance spacing ( $L_2-L_1$ ) = $\text{20}\text{.2 cm}$

To calculate wavelength use formula

  $\frac{\lambda }{2}=L_2-L_1$

Substitute the value ( $L_2-L_1$ ) = $\text{20}\text{.2 cm}$

  $\begin{array}{l}\frac{\lambda }{2}=20.2\text{ cm}\\ \lambda =2(20.2\text{ cm)}\\ \lambda =40.4\text{ cm}\\ \lambda \text{= 0}\text{.404 m}\end{array}$

To calculate frequency of the tuning fork use formula

  $f=\frac{v}{\lambda }$

Substitute the values $v=347\text{ m/s}$ and $\lambda \text{= 0}\text{.404 m}$

  $f=\frac{347\text{ m/s}}{0.404\text{ m}}=859\text{ Hz}$

Conclusion:

Thus, the frequency of the tuning fork is $859\text{ Hz}$ .