Chapter 12, Problem 76GP

To determine

To Find: The frequency of tuning fork.

Answer to Problem 76GP

The frequency of the tuning fork is $f=635.5\text{ Hz}$

Explanation of Solution

Given:

Distance from the tube opening = 0.125 and 0.395,

Formula used:

  $f=\frac{v}{\lambda }\text{ Hz}$

Here, $v$ is the velocity,

  $\lambda$ is the wavelength of tuning fork.

Calculation:

The distance between two resonances is distance between two adjacent nodes.

Therefore, the wavelength ? can be calculated as,

  $\frac{\lambda }{2}=0.395-0.125$

  $\lambda =2\times 0.270$

  $\lambda =0.54\text{ m}$

Now, calculate the frequency ( f ),

  $f=\frac{v}{\lambda }\text{ Hz}$

  $f=\frac{343\text{ m/s}}{0.54\text{ m}}$

  $f=635.5\text{ Hz}$

Conclusion:

The frequency of tuning fork is 635.5 Hz.