Question

$A_1$ and $A_2$ are the two amplitudes of the two waves producing beats. Now when $A_1=A_2$

• A. $I_{max}=4A^2$ and $I_{min}=2A^2$
• B. $I_{max}=4A^2$ and $I_{min}=0$
• C. $I_{max}=2A^2$ and $I_{min}=0$
• D. $I_{max}=16A^2$ and $I_{min}=4A^2$

Answer 1

Answer: (B)

$I_{max}=4A^2$ and $I_{min}=0$

Maximum resultant amplitude of the wave is given by $I_{max}=(A_1+A_2{)}^2$
Given : $A_1=A_2=A$
$\therefore$ $I_{max}=(A+A{)}^2=(2A{)}^2=4A^2$
Minimum resultant amplitude of the wave is given by $I_{min}=(A_1-A_2{)}^2$
$\therefore$ $I_{min}=(A-A{)}^2=0$