The wavelengths of two sound notes in air are $\frac{40}{195} \mathrm{~m}$ and $\frac{40}{193} \mathrm{~m}$.…
The wavelengths of two sound notes in air are $\frac{40}{195} \mathrm{~m}$ and $\frac{40}{193} \mathrm{~m}$. Each note produces 9 beats per second separately with a third note of fixed frequency. The velocity of sound in air in $\mathrm{m} / \mathrm{s}$ is
360
320
300
340
Solution
Here, $\quad v_1=\frac{v}{\left(\frac{40}{195}\right)}$
$\begin{array}{ll}\text { and } & v_2=\frac{v}{\left(\frac{40}{193}\right)} \\ \Rightarrow & v_1=\frac{v(195)}{40}\end{array}$
and $\quad v_2=\frac{v(193)}{40}$
According to given condition
Adding Eqs. (i) and (ii)
$\begin{aligned}
18=\frac{v}{40}(195-193) & =\frac{2 v}{40} \Rightarrow 18=\frac{v}{20} \\
\text { or } \quad v & =360 \mathrm{~m} / \mathrm{s}
\end{aligned}$
