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

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(B) The resultant amplitude $A$ of two interfering waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:$A = \sqr

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$6$

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(B) The resultant amplitude $A$ of two interfering waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:$A = \sqrt{a_1^2 + a_2^2 + 2 a_1 a_2 \cos \phi}$Here,$a_1 = 4$,$a_2 = 3$,and $\phi = \frac{\pi}{3} = 60^{\circ}$.Substituting these values into the formula:$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos 60^{\circ}}$$A = \sqrt{16 + 9 + 24 	imes 0.5}$$A = \sqrt{25 + 12}$$A = \sqrt{37} \approx 6.08$Thus,the amplitude of the resulting wave is approximately $6$.

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

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