Chapter 15.2, Problem 13PP

To determine

To calculate:The velocity of sound in helium gas when a tuning fork is used with a resonating column.

Answer to Problem 13PP

Thevelocity of sound in helium gas is $968\text{ m/s}$ .

Explanation of Solution

Given info:

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

Resonance spacing = $\text{110 cm}$

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,

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

Resonance spacing ( $L_2-L_1$ ) = $\text{110 cm}$

To calculate wavelength use formula

  $\frac{\lambda }{2}=L_2-L_1$

Substitute the value $L_2-L_1=110\text{ cm}$ ,

  $\begin{array}{l}\frac{\lambda }{2}=110\text{ cm}\\ \lambda =2(110\text{ cm)}\\ \lambda =220\text{ cm}\\ \lambda \text{= 2}\text{.20 m}\end{array}$

To calculate velocity of sound in helium rearrange the formula

  $f=\frac{v}{\lambda }$

  $v=f\lambda$

Substitute the values $f=440\text{ Hz}$ and $\lambda =2.20\text{ m}$

  $\begin{array}{l}v=(440\text{ Hz)}(2.20\text{ m)}\\ v=968\text{ m/s}\end{array}$

Conclusion:

Thus, thevelocity of sound in helium gas is $968\text{ m/s}$ .