Q. Two sound waves with wavelengths $5.0\ m$ and $5.5\ m$ respectively, each propagate in a gas with velocity $330\ m/s$ We expect the following number of beats per second
A) 1
B) 6
C) 12
D) 0
Explanation
(b) $Frequency = \frac{velocity}{Wavelength}$
$\therefore f_{1} = \frac{v}{\lambda_{1}} = \frac{330}{5} = 66Hz$
And $f_{2} = \frac{v}{\lambda_{2}} = \frac{330}{5.5} = 60Hz$
Number of beats per second
$= f_{1} - f_{2} = 66 - 60 = 6$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.