Posted by sarah on .
A tuning fork is set into vibration above a vertical open tube filled with water. 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 0.145 m, and again at 0.345 m. What is the frequency (in hertz) of the tuning fork?
We have the standing sound wave in a pipe; an open end is the antinode for air column, a closed end (the water surface) is the node.
The distance between two adjacent nodes is the half of the wavelength:
λ/2 = 0.345 – 0.145 = 0.2 m, λ = 0.4 m.
λ = v•T = v/f.
In dry air at 20 °C the speed of sound is v = 343.2 m/s.
f =v/ λ = 343.2/0.4 = 858 Hz