Question

The displacement at a point due to two wave are $y_1=4sin\left(500\pi t\right)$ and $y_2=2sin\left(506\pi t\right)$ 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 Betas per seconds 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)

3 Betas per seconds with intensity relation between maxima and minima equal to 9

Since, the general equation of wave is $y=A\sin \left(2\pi ft\right)$

Therefore,

$A_1=4,f_1=250$

$A_2=2,f_2=253$

So, beats per second $=|f_1-f_2|=3$

Now,

Maximum Amplitude $=(A_1+A_2{)}^2=36$

Minimum Amplitude $=(A_1-A_2{)}^2=4$

Intensity relation of maxima and minima $=\frac{36}{4}=9$

So, the correct option will be $\left(B\right)$.