A Kundt's tube apparatus has a steel rod of length it 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.

Open in App

Solution

Here given, $L = \frac{1.0}{2} = 0.5 m$

$d_{3} = 6.5 c m = 6.5 \times 10^{- 2} m$

As Kundt's tube Apparatus is a closed organ pipe, its fundamental frequency,

$\Rightarrow n = \frac{1.0}{2}$

$\Rightarrow V_{r} = 2600 \times 4 \times 0.5$

r =5200 m/s

$(b) \frac{V_{r}}{V_{a}} = \frac{2 L_{r}}{d_{a}}$

$\Rightarrow V_{r} = \frac{5200 \times 6.5 \times 10^{- 2}}{2 \times 0.5}$

=338 m/s

Suggest Corrections

Similar questions

A Kundt's tube apparatus has a copper rod of length 1.0 m clamped at 25 cm from one of the ends. The tube contains air in which the speed of sound is 340 in/s. The powder collects in heaps separated by a distance of 5.0 cm. Find the speed of sound waves in copper.

Q. In a Kundt's tube experiment, the heaps of lycopodium powder are collected at

20 c m

separations. The frequency of tuning fork used is

Join BYJU'S Learning Program

A Kundt's tube apparatus has a steel rod of length it 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.

Here given, L=1.02=0.5m d3=6.5 cm=6.5×10−2m As Kundt's tube Apparatus is a closed organ pipe, its fundamental frequency, ⇒n=1.02 ⇒Vr=2600×4×0.5 r =5200 m/s (b)VrVa=2Lrda ⇒Vr=5200×6.5×10−22×0.5 =338 m/s