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 $y_{1}\left(f_{1}\right)=\frac{\omega}{2 \pi}$
$=\frac{500 \pi}{2 \pi}=250\, Hz$
Frequency of $y_{2}\left(f_{2}\right)=\frac{\omega}{2 \pi}$
$=\frac{506 \pi}{2 \pi}=253 \,Hz$
Intensity relation $\frac{A_{\max }}{A_{\min }}=\frac{\left(A_{1}+A_{2}\right)^{2}}{\left(A_{1}-A_{2}\right)^{2}}$
$=\frac{36}{4}=9$