The displacement at a point due to two waves | vedclass

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(B) Given wave equations are $y_1 = 4 \sin(500 \pi t)$ and $y_2 = 2 \sin(506 \pi t)$.Comparing with $y = A \sin(2 \pi f t)$,we get frequencies $f_1 =

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$3$ beats per second with intensity relation between maxima and minima equal to $9$

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(B) Given wave equations are $y_1 = 4 \sin(500 \pi t)$ and $y_2 = 2 \sin(506 \pi t)$.Comparing with $y = A \sin(2 \pi f t)$,we get frequencies $f_1 = \frac{500 \pi}{2 \pi} = 250 \, Hz$ and $f_2 = \frac{506 \pi}{2 \pi} = 253 \, Hz$.Beat frequency is given by $f_b = |f_2 - f_1| = |253 - 250| = 3 \, Hz$.This means $3$ beats per second are produced.The amplitudes are $A_1 = 4$ and $A_2 = 2$.The intensity $I$ is proportional to the square of the amplitude $(I \propto A^2)$.The ratio of maximum intensity to minimum intensity is given by $\frac{I_{\max}}{I_{\min}} = \frac{(A_1 + A_2)^2}{(A_1 - A_2)^2}$.Substituting the values: $\frac{I_{\max}}{I_{\min}} = \frac{(4 + 2)^2}{(4 - 2)^2} = \frac{6^2}{2^2} = \frac{36}{4} = 9$.Thus,the result is $3$ beats per second with an intensity ratio of $9$.

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

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