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 Hz is held just over the open top end of the tube . At what positions of the water level will there be resonance . Speed of sound is 330 m//s.

The water level in a vertical glass tube 1.0 m long can be adjusted to

1.	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 Hz is held just over the open top end of the tube . At what positions of the water level will there be resonance . Speed of sound is 330 m//s.
Answer» Solution :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 antinodesis `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.