The displacements of two interfering light wa | vedclass

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(A) The given equations for the two waves are $y_1 = a_1 \sin \omega t$ and $y_2 = a_2 \sin (\omega t + \phi)$.Comparing these with the given equation

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$5$

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(A) The given equations for the two waves are $y_1 = a_1 \sin \omega t$ and $y_2 = a_2 \sin (\omega t + \phi)$.Comparing these with the given equations,we have $a_1 = 4$,$a_2 = 3$,and the phase difference $\phi = \frac{\pi}{2}$.The amplitude $A$ of the resultant wave is given by the formula $A = \sqrt{a_1^2 + a_2^2 + 2a_1 a_2 \cos \phi}$.Since $\cos(\frac{\pi}{2}) = 0$,the formula simplifies to $A = \sqrt{a_1^2 + a_2^2}$.Substituting the values,$A = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$.

The displacements of two interfering light waves are given by $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \frac{\pi}{2})$. What is the amplitude of the resultant wave?

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