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(B) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:$A = \sqrt{a_1^2 + a_

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$6$

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(B) The resultant amplitude $A$ of two waves with amplitudes $a_1$ and $a_2$ and phase difference $\phi$ is given by the formula:$A = \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \cos \phi}$Given $a_1 = 4$,$a_2 = 3$,and $\phi = \frac{\pi}{3}$.Substituting these values into the formula:$A = \sqrt{4^2 + 3^2 + 2(4)(3) \cos(\frac{\pi}{3})}$Since $\cos(\frac{\pi}{3}) = 0.5$:$A = \sqrt{16 + 9 + 24(0.5)}$$A = \sqrt{25 + 12}$$A = \sqrt{37}$$A \approx 6.08$Rounding to the nearest integer,the resultant amplitude is $6$.

The resultant amplitude of the interference of two waves $y_1 = 4 \sin \omega t$ and $y_2 = 3 \sin (\omega t + \frac{\pi}{3})$ is:

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