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

Given:

• Length of tube = 1 m

• Frequency, f = 600 Hz
• Speed of sound , v= 300 m/s

To find:

• Positions of the water level  which will be there in resonance

Answer:

• Resonance occurs at pressure anti-node at closed end and pressure node at open end .
• Distance between anti-node and node = λ/4 = v/(4f)
• Length of air column = n * Distance between anti-node and node

                                           = n (v/4f) =  $n\frac{300}{4*600} = \frac{n}{8}$   , where n=1,3,5,...

• L₁ = $\frac{1}{8} = 0.125 m$

        L₂ = 3 L₁ = 0.375 m

        L₃ = 5 L₁ = 0.625 m

        L₄ = 7 L₁ = 0.875 m

        L₅ = 9 L₁ = 1.125 m

• But L₅ > 1 m , so only rest four are considered.

• So, 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.