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If two waves represented by $y_1=4 \sin \omega t$ and $y_2=3 \sin \left(\omega t+\frac{\pi}{3}\right)$ interfere at a point, the amplitude of the resulting wave will be about
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The correct answer is:
$6$
$\phi=\pi / 3, a_1=4, a_2=3$
So, $A=\sqrt{a_1^2+a_2^2+2 a_1 \cdot a_2 \cos \phi} \Rightarrow A \approx 6$
So, $A=\sqrt{a_1^2+a_2^2+2 a_1 \cdot a_2 \cos \phi} \Rightarrow A \approx 6$
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