Question

Two waves of wavelengths 99 cm and 100 cm both travelling with velocity 396 m/s are made to interfere. The number of beats produced by them per second is

• A. 1
• B. 2
• C. 4
• D. 8

Answer 1

The correct option is C 4

Given : ${\lambda }_1=99cm=0.99m$ and ${\lambda }_2=100cm=1m$

Velocity of wave $v=396m/s$

Frequency of first wave $f_1=\frac{v}{{\lambda }_1}=\frac{396}{0.99}$

Frequency of second wave $f_2=\frac{v}{{\lambda }_2}=\frac{396}{1}$

Thus number of beat produced per second $b=f_1-f_2=396[\frac{1}{0.99}-\frac{1}{1}]$

$\therefore$ $b=4$