Question 31

The amplitude of the vibrating particle due to superposition of two SHMs

$$x_1=\ 2\ Sin\left(\omega t\right)$$ and

$$x_2=\ 2\ Sin\left(\omega t+\ \frac{2\pi}{3}\right)$$ .

The resultant amplitude of two SHMs having amplitudes $$A_1$$ and $$A_2$$ with phase difference $$\phi$$ is

$$A=\sqrt{A_1^2+A_2^2+2A_1A_2\cos\phi}$$

Here,

$$A_1=A_2=2,\qquad \phi=\frac{2\pi}{3}$$

Therefore, $$A=\sqrt{2^2+2^2+2(2)(2)\cos\frac{2\pi}{3}}$$

$$A=\sqrt{4+4+8\left(-\frac{1}{2}\right)}$$

$$A=\sqrt{4}=2$$

Therefore, $${A=2}$$

Hence, the correct option is B.

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