Question

The wavelengths of two sound notes in air are $\frac{40}{195}m$ and $\frac{40}{193}m$. Each note produces $9$ beats per second separately with a third note of fixed frequency. The velocity of sound in air in $m/s$ is:

• A. $360$
• B. $320$
• C. $300$
• D. $340$

Answer 1

The correct option is A $360$

Let the speed of sound in air be $v$ and the fixed frequency of the third note be $\nu$.

Given that:

${\nu }_1=\frac{v}{\frac{40}{195}}=\frac{195v}{40}$

and, ${\nu }_2=\frac{v}{\frac{40}{193}}=\frac{193v}{40}$

According to the given conditions:

$\frac{195v}{40}-\nu =9$ ......... (I)

and, $\nu -\frac{193v}{40}=9$ ......... (II)

Adding equations (I) and (II), we get:

$18=\frac{v}{40}(195-193)$

$\Rightarrow 18=\frac{v}{20}$

$\Rightarrow v=360m/s$