Question

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 0.125 m, and again at 0.385 m. What is the frequency (in hertz) of the tuning fork?

Answer 1

Answer

given,

speed of sound = 343 m/s

distance of tube opening to water level = 0.125 m

and another opening at a distance of = 0.385 m

frequency of tuning fork = ?

The difference between two different modes is given by:

$L_{n+2} - L_{n} = \dfrac{(n+2)l}{4}-\dfrac{nl}{4}$

$L_{n+2} - L_{n} = \dfrac{l}{2}$

$0.385 - 0.125 = \dfrac{l}{2}$

$0.26 = \dfrac{l}{2}$

l = 0.26 x 2

l = 0.52 m

$f = \dfrac{v}{l}$

$f = \dfrac{343}{0.52}$

f = 659.6 Hz