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

12

0

1

6

Solution

Let $straight lambda subscript 1 space equals space 5.0 space straight m comma space space straight v space equals space 330 space straight m divided by straight s space and space straight lambda subscript 2 space equals space 5.5 space straight m$
The relation between frequency , wavelength and velocity is given by
$straight v equals nλ$
   $rightwards double arrow space space space space straight n space equals space straight v over straight lambda$ ...(i)
The frequency corresponding to wavelength, $straight lambda subscript 1$
   $straight n subscript 1 space equals space straight v over straight lambda subscript 1 space equals space fraction numerator 330 over denominator 5.0 end fraction space equals space 66 space Hz$
The frequency corresponding to wavelength $straight lambda subscript 2 comma$
   $straight n subscript 2 space equals straight v over straight lambda subscript 2 space equals space fraction numerator 330 over denominator 5.5 end fraction space equals space 60 space Hz$
Hence, no of beats per second
   $equals straight n subscript 1 minus straight n subscript 2 equals 66 minus 60 equals space space 6$