Question

Two sound waves of wavelengths ${\lambda }_1$ and ${\lambda }_2$ $({\lambda }_2>{\lambda }_1)$ produce $n$ beats per second. Then, the speed of sound is

• A. $\frac{n{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}$
• B. $n({\lambda }_1+{\lambda }_2)$
• C. $n({\lambda }_2-{\lambda }_1)$
• D. $n(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

Answer 1

The correct option is A $\frac{n{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}$

From formula,
beat frequency $=f_1-f_2(\because {\lambda }_2>{\lambda }_1)$
$\Rightarrow n=f_1-f_2$
$\Rightarrow n=\frac{v}{{\lambda }_1}-\frac{v}{{\lambda }_2}[\because v=\lambda f]$
$\Rightarrow n=v[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}]$
$\Rightarrow v=\frac{n{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}$
So, speed of sound is $\frac{n{\lambda }_1{\lambda }_2}{{\lambda }_2-{\lambda }_1}$