Wavelengths of two notes in air are (\frac{90}{175}) m and (\frac{90}{173}) m. Each note produces 4 beats/s…
Wavelengths of two notes in air are (\frac{90}{175}) m and (\frac{90}{173}) m. Each note produces 4 beats/s with a third note of a fixed frequency. Calculate the velocity of sound in air.
Solution
Sol. Given, $\lambda_1 = \frac{90}{175}\ \mathrm{m}$ and $\lambda_2 = \frac{90}{173}\ \mathrm{m}$.
Let $f_1$ and $f_2$ be the corresponding frequencies and $v$ be the velocity of sound in air.
$v = \lambda_1 f_1$ and $v = \lambda_2 f_2$
∴ $f_1 = \frac{v}{\lambda_1}$ and $f_2 = \frac{v}{\lambda_2}$
Since, $\lambda_2 > \lambda_1$
∴ $f_1 > f_2$
Let $f$ be the frequency of the third note.
∴ $f_1 - f = 4$ and $f - f_2 = 4$
∴ $f_1 - f_2 = 8$
∴ $\displaystyle \frac{v}{\lambda_1} - \frac{v}{\lambda_2} = 8$
⇒ $v\left(\frac{175}{90} - \frac{173}{90}\right) = 8$
⇒ $v\left(\frac{2}{90}\right) = 8$
Therefore, $v = 360\ \mathrm{ms^{-1}}$
Answer: $360\ \mathrm{ms^{-1}}$