Question

36. The water level in a vertical glass tube 1m long can be adjusted to any position in the tube. A tuning fork vibrating at 520 Hz is held just over the open top end of the tube. At what position of the water level will there be resonance? ( velocity of sound = 330m/s).

Answer 1

Answer:

The frequency of the tuning fork is 500 Hz.

Given data:

Velocity of the sound = 340 m/s

Resonances observed at two successive lengths, l₁=0.50 m and l₂=0.84 m.

To find: Frequency of the tuning fork

In the above question we are give two successive lengths therefore from the equation “(2n+1)λ/4” we get the successive lengths as λ/4 and 3λ/4.

Therefore,

For the first resonance, l₁ + e = λ/4 ….. (i)

For the second resonance, l₂ + e = 3λ/4 … (ii)

Subtracting equation (i) from (ii), we get

l₂ – l₁ = 3λ/4 - λ/4 = 2λ/4 = λ/2

or, λ/2 = l₂ – l₁

or, λ = 2 (0.84 – 0.50) = 0.68 m

velocity of the sound(v) = frequency(f) * wavelength(λ)

or, f = v/ λ

or, f = 340 / 0.68 = 500 Hz

Explanation:

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