If two waves represented by y_1 = 4sin omega | vedclass

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(B) The given wave equations are $y_1 = 4\sin \omega t$ and $y_2 = 3\sin (\omega t + \pi/3)$.Comparing these with the standard form $y = a\sin(\omega

Vedclass · Accepted Answer

$6$

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(B) The given wave equations are $y_1 = 4\sin \omega t$ and $y_2 = 3\sin (\omega t + \pi/3)$.Comparing these with the standard form $y = a\sin(\omega t + \phi)$,we get amplitudes $a_1 = 4$ and $a_2 = 3$,and the phase difference $\phi = \pi/3$.The amplitude $A$ of the resulting wave is given by the formula $A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$.Substituting the values: $A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\pi/3)}$.Since $\cos(\pi/3) = 0.5$,we have $A = \sqrt{16 + 9 + 24(0.5)} = \sqrt{25 + 12} = \sqrt{37}$.Calculating the square root,$A \approx 6.08$,which is approximately $6$.

If two waves represented by $y_1 = 4\sin \omega t$ and $y_2 = 3\sin (\omega t + \pi/3)$ interfere at a point,the amplitude of the resulting wave will be about

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