Question

A sound tube, which measures 50 cm, is closed AT BOTH ENDS and has a reed that produces vibrations in the middle. The air is at 0°C and the speed of sound is c= 340 m/s.
a) What is the frequency of the tub?
c) How much should the air be cooled in order to obtain the next harmonic?

Answer 1

Answer:

The frequency of vibration of air column in th pipe is f1=3v4l=3×3404×0.85=300Hz.

The frequency of vibration of string is f2=f1+6=300+6=306Hz

Now f2=32lFμ−−√

306=32×0.5F×0.52.5×10−3−−−−−−−−−√

306=302–√F

sqrt2F = 10.2 :. F= 52 N`.