The water level in a vertical glass tube `1.0 m` long can be adjusted to any position in the tube . `A` tuning fork vibrating at `660 H_(Z)` is held just over the open top end of the tube . At what positions of the water level wil they be in resonance? Speed of sound is `330 m//s` .

Explanation

Resonance corresponds to a pressure antinode at closed end and pressure node at open end. Further , the distance between a pressure node and a pressure antinodes is `lambda//4` , the condition of resonance would be ,
length of air column
` l = n (lambda)/(4) = n((v)/( 4 f))`
Here , ` n = 1 , 3, 5 ,....`
`l_(1) = (1) (( 330)/( 4 xx 660)) = 0.125 m`
and `l_(2) = 3 l_(1) = 0.125 m`
`l_(3) = 5l_(1) = 0.625 m`
and `l_(4) = 7 l_(1) = 0.875 m`
`l_(3) = 9l_(1) = 1.125 m`
Since `l_(5) gt m` ( the length of tube), the length of air columns can have the values from `l_(1) to l_(4)` only. Therefore , level of water at resonance will be
`( 1.0 - 0.125 ) m = 0.875 m`
`( 1.0 - 0.375)m = 0.625 m`
`( 1.0 - 0.625)m = 0.375 m`
and `( 1.0 - 0.875) m = 0.125 m`
In all the four cases shown in Fig . 7.68, the resonance frequency is `660 Hz` but the first one is the fundamental tone or first harmonic . The second is first overtone or the third harmonic and so on.