Chapter 15.2, Problem 15PP

To determine

To calculate:The spacing between the resonances.

Answer to Problem 15PP

The spacing between the resonances is $39\text{ cm}$ .

Explanation of Solution

Given info:

Speed of sound in air at ${20}^{\text{o}}\text{C}$ = $343\text{ m/s}$

Frequency of the tuning fork = $440\text{ Hz}$

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.

And the frequency- wavelength relation 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 ${20}^{\text{o}}\text{C}$ ( $v$ ) = $343\text{ m/s}$

Frequency of the tuning fork ( $f$ ) = $440\text{ Hz}$

To calculate wavelength rearrange the formula

  $\begin{array}{l}f=\frac{v}{\lambda }\\ \lambda =\frac{v}{f}\end{array}$

Substitute the values $v=343\text{ m/s}$ and $f\text{= 440 Hz}$

  $\lambda =\frac{343\text{ m/s}}{440\text{ Hz}}=0.78\text{ m}$

To calculate resonance spacing use formula

  $\frac{\lambda }{2}=L_2-L_1$

Substitute the value $\lambda \text{= 0}\text{.78 m}$

  $\begin{array}{l}{L}_2-L_1=\frac{0.78\text{ m}}{2}\\ L_2-L_1=0.39\text{ m}\\ L_2-L_1=39\text{ cm}\end{array}$

Conclusion:

Thus, thespacing between the resonances is $39\text{ cm}$ .