Question

A closed organ pipe gives first resonance with a tuning fork of frequency 500 Hz
when the length of air column is 15 cm. It gives a second resonance with the same
tuning fork when the length of air column is 45 cm. Calculate the velocity of sound in
air.
If the prong of tuning fork is loaded with a little wax, how will its frequency change?
Explain how will the velocity of sound in air change with the increase in the
(1) Temperature (ii) Humidity ? Explain.
​

Answer 1

Answer:

720ms.............. .

Answer 2

In closed organ pipe for fundamental node

of vibration,

λ/4=0.2⇒λ=0.8m,

v=nλ

If we take the frequency of air column to be equal to tuning fork then,

v=450×0.8=360m/sec

If we take the frequency of air column to be twice that of tuning fork then

v=900×0.8=720m/sec.