Looking for classes? Ksquare Career Institute, Bengaluru →
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
12
B
0
C
1
D
6
Detailed Solution
Frequency of a wave: $n = \dfrac{v}{\lambda}$
$n_1 = \dfrac{330}{5.0} = 66$ Hz
$n_2 = \dfrac{330}{5.5} = 60$ Hz
Number of beats per second $= n_1 - n_2 = \dfrac{v}{\lambda_1} - \dfrac{v}{\lambda_2} = 330\left(\dfrac{1}{5} - \dfrac{1}{5.5}\right)$
$= 66 - 60$
$= 6$
$n_1 = \dfrac{330}{5.0} = 66$ Hz
$n_2 = \dfrac{330}{5.5} = 60$ Hz
Number of beats per second $= n_1 - n_2 = \dfrac{v}{\lambda_1} - \dfrac{v}{\lambda_2} = 330\left(\dfrac{1}{5} - \dfrac{1}{5.5}\right)$
$= 66 - 60$
$= 6$