Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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# 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. - Physics

Sum

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.

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#### Solution

Given :
Length at which steel rod is clamped l = $\frac{1}{2} = 0 . 5 \text { m }$

Fundamental mode of frequency f = 2600 Hz
Distance between the two heaps $∆ l$ = 6.5 cm = $6 . 5 \times {10}^{- 2} \text { m }$

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

$f = \frac{v_{air}}{4L}$

$\Rightarrow v_{air} = f \times 2 \times ∆ L$

$\Rightarrow v_{air} = 2600 \times 2 \times 6 . 5 \times {10}^{- 2} = 338 \text { m/s }$

$(b) \frac{v_{steel}}{v_{air}} = \frac{2 \times l}{∆ l}$

$\Rightarrow v_{steel} = \frac{2l}{∆ l} \times v_{air}$

$\Rightarrow v_{steel} = \frac{2 \times 0 . 5 \times 338}{6 . 5 \times {10}^{- 2}}$

$\Rightarrow v_{steel} = 5200 \text { m/s }$

Concept: Speed of Wave Motion
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 55 | Page 356
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