Question

A long glass tube is held vertically in water. A tuning fork is struck
and held over the tube. Strong resonances are observed at two
successive lengths 0.50 m and 0.84 m above the surface of water. If
the velocity of sound is 340 m/s, then the frequency of the tuning
fork is​

Answer 1

Answer:

The frequency of the tuning fork is 500 Hz.

Explanation:

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