Question

Question: A tuning fork is set into vibration above a vertical open tube filled with water (Fig 12-40). The water level is allowed to drop slowly. As it does so, the air in the tube above the water level is heard to resonate with the tuning fork when the distance from the tube opening to the water level is 0.125 m and again at 0.395 m. What is the frequency of the tuning fork?

FIGURE 12–40 Problem 79.

Short answer

The frequency of the tuning fork is \(635\;{\rm{Hz}}\).

Step-by-step solution

Determination of frequency of tuning fork

The distance between the two nodes is half of the wavelength. Using this concept, the wavelength is determined, hence the tuning fork's frequency.

Given information

The length of the tube is 0.395 m.

The water level is 0.125 m below the top.

Find the wavelength of the tuning fork

The distance between the node is one-half of the length. Therefore,

\(\begin{array}{c}\Delta l = \frac{1}{2}\lambda \\\lambda = 2\Delta l\\ = 2\left( {0.395\,\,{\rm{m}} - 0.125\,\,{\rm{m}}} \right)\\ = 0.540\;{\rm{m}}\end{array}\)

Find the frequency of the tuning fork

The value of frequency can be calculated as:

Thus, the frequency of the tuning fork is \(635\;{\rm{Hz}}\).