(c) $y_{1} = 4\sin{\omega t}$ $y_{2} = 3\sin(\omega t + \pi/3)$ Here, $a = 4,b = 3,\phi = \pi/3$ $R = \sqrt{a^{2} + b^{2} + 2ab\cos\phi}$ $= \sqrt{4^{2} + 3^{2} + 2 \times 4 \times 3\cos{\pi/3}}$ $= \sqrt{37} = 6$
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.