Question

A tuning fork causes resonance in a closed pipe. The difference between the length of the closed tube for the first resonance and the second resonance is 54.0 cm. If the tuning forks frequency is 320 Hz, find the wavelength and speed of the sound waves.

Answer 1

Given data:

* The difference between the first and second resonance is d = 54 cm.

* The frequency of the tuning fork is f = 320 Hz.

Solution:

The length of the pipe for the first resonance is,

`L_1=(\lambda)/(4)`

The length of pipe for the second resonance is,

`L_2=(3\lambda)/(4)`

The difference in the length for first and second resonance is,

`\begin{gathered} L_2-L_1=(3\lambda)/(4)-(\lambda)/(4)_{} \\ L_2-L_1=(2\lambda)/(4) \\ L_2-L_1=(\lambda)/(2) \\ d=(\lambda)/(2) \end{gathered}`

Substituting the known values,

`\begin{gathered} 54=(\lambda)/(2) \\ \lambda=108\text{ cm} \\ \lambda=1.08\text{ m} \end{gathered}`

Thus, the wavelength of a sound wave is 1.08 m.

The speed of the sound wave is,

`\begin{gathered} v=f\lambda \\ v=320*1.08 \\ v=345.6\text{ m/s} \end{gathered}`

Thus, the speed of the sound wave is 345.6 m/s.