Question

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 ther be in resonance? Speed of sound is `330 m//s` .

Answer 1

Correct Answer - A::B
Resonance corresponds to a pressure antinode at closed end and pressure node at open end . Further, the distance between a pressure node a pressure antinode is `(lambda) / (4)` , the condition of resonance would be,
Length of air column `l = n (lambda)/(4) =n ((upsilon)/(4f))`
Here, `n = 1,3,5,...`
`l_(1)= (1) ((330)/(4xx660))= 0.125 m`
`l_(2) = 3l_(1)= 0.375 m`
`l_(3)= 5l_(1) = 0.625 m`
`l_(4) = (7l_(1) = 0.875 m`
`l_(5)= 9l_(1)= 1.125 m`
Since, `l_(5)gt 1 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 figure, the resonance frequency in `660 H_(Z)` but first one is the fundamental tone or first harmonic. Second is first overtone or third harmonic and soon.