Exercise 10.13
The variation of density of medium for a sound wave propagating with a speed of m is shown in Fig. 10.32. Calculate the wavelength and frequency of the sound wave.
NCERT’s answer
0.$\displaystyle 04$ m; $\displaystyle 8500$ Hz
Wavelength \(\displaystyle \lambda = 0.04\ \text{m} \) ($\displaystyle 4$ cm) and frequency \(\displaystyle \nu = 8500\ \text{Hz} \).
The $\displaystyle 8$ cm marked in Fig. $\displaystyle 10.32$ spans two complete cycles — two compressions and two rarefactions — so one wavelength is \(\displaystyle \lambda = \dfrac{8\ \text{cm}}{2} = 4\ \text{cm} = 0.04\ \text{m} \).
From \(\displaystyle v = \nu \times \lambda \), \(\displaystyle \nu = \dfrac{v}{\lambda} = \dfrac{340\ \text{m s}^{-1}}{0.04\ \text{m}} = 8500\ \text{Hz} \).
\(\displaystyle 8500\ \text{Hz} \) lies inside the audible range of $\displaystyle 20$ Hz to $\displaystyle 20$ kHz, so this sound can be heard.