Question

The displacement at a point due to two waves are $y_1=4\sin (500\pi t)$ and $y_2=2\sin (506\pi t)$. The result due to their superposition will be

• A. 3 beats per second with intensity relation between maxima and minima equal to 2
• B. 3 beats per second with intensity relation between maxima and minima equal to 9
• C. 6 beats per second with intensity relation between maxima and minima equal to 2
• D. 6 beats per second with intensity relation between maxima and minima equal to 9

Answer 1

Answer: (B)

$y_1=4\sin 500\pi t$
$y_2=2\sin 506\pi t$
Frequency of $y_1\left(f_1\right)=\frac{\omega }{2\pi }$
$=\frac{500\pi }{2\pi }=250Hz$
Frequency of $y_2\left(f_2\right)=\frac{\omega }{2\pi }$
$=\frac{506\pi }{2\pi }=253Hz$
Intensity relation $\frac{A_{\max }}{A_{\min }}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1-A_2\right)}^2}$
$=\frac{36}{4}=9$