Question

A Kundt's tube apparatus has a steel rod of length 1.0 m clamped at the centre. It is vibrated in its fundamental mode at a frequency of 2600 Hz. The lycopodium powder dispersed in the tube collects into heaps separated by 6.5 cm. Calculate the speed of sound in steel and in air.

Answer 1

Given:

Length at which steel rod is clamped l = 12=0.5 m

Fundamental mode of frequency f = 2600 Hz

Distance between the two heaps ∆l = 6.5 cm = 6.5×10−2 m

Since Kundt's tube apparatus is a closed organ pipe, its fundamental frequency is given by:

f=vair4L⇒vair=f×2×∆L⇒vair=2600×2×6.5×10−2=338 m/s(b) vsteelvair=2×l∆l⇒vsteel =2l∆l×vair⇒ vsteel=2×0.5×3386.5×10−2⇒ vsteel=5200 m/s