A tuning fork is set into vibration above a vertical open tube filled with water

Question

A tuning fork is set into vibration above a vertical open tube filled with water (the figure). 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 d1 = 0.128m and again at d2 = 0.394m .What is the frequency of the tuning fork?

A tuning fork is set into vibration above a vertical open tube filled with water (the figure). 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 d1= 0.128m and again at d2= 0.394m . (Figure 1) What is the frequency of the tuning fork?

Explanation / Answer

The wavelength corresponding to the resonant frequency be w, then

w/4 = (.125+e), where e is the excess length.

3w/4 = (.395+e)

Subtract the two eqs. to get w = 0.54 metres.

The speed of the sound isn't given. I will take it to be 330 m/s

frequency of the tuning fork = 330 / .54 = 611.11 Hz

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