Physics - SHM Question with Solution | TestHub

Given, $x_1=A_1\sin \left(\omega t+{\varphi }_1\right)$ and $x_2=A_2\sin \left(\omega t+{\varphi }_2\right)$ A: The two SHMs are in phase, $A_1=A_2=A$ $R=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \delta }=A\sqrt{2(1+\cos \delta )}=2A\cos \left(\frac{\delta }{2}\right)$ where, the phase difference is $\delta =0$. $R=2A$ Therefore, A - III. $\mathrm{B}$ :The two SHMs are in phase, $A_1\ne A_2$ and phase difference $\delta =0$. $\begin{array}{rl}& R=\sqrt{A_1{}^2+A_2{}^2+2A_1{A}_2\cos \delta }\text{ and }\cos \delta =\cos 0^{\circ }=1\\ & \Rightarrow R=\sqrt{A_1^2+A_2^2+2A_1{A}_2}=\left(A_1+A_2\right)\end{array}$ Therefore, B - I. C: The two SHMs are $90^{\circ }$ out of phase, $A_1=A_2=A$ and phase difference $\delta =90^{\circ }$. $R=\sqrt{A_1^2+A_2^2+2A_1{A}_2\cos \delta }=\sqrt{A^2+A^2}=A\sqrt{2}$ Therefore, C - IV. $\mathrm{D}$ :The two SHMs are $180^{\circ }$ out of phase, $A_1=A_2$ and phase difference $\delta =180^{\circ }$. $R=\sqrt{A_1{}^2+A_2{}^2+2A_1{A}_2\cos \delta }=\sqrt{A^2+A^2-2A^2}=0$ Therefore, D - II.