Consider two SHMs along the same straight line \(x_1=A_1 \sin \left(\omega t+\phi_1\right)\), \(x_2=A_2 \sin \left(\omega t+\phi_2\right)\), where \(A_1\) and \(A_2\) are their amplitudes and \(\phi_1\) and \(\phi_2\) are their initial phase angle. If the two SHMs meet simultaneously and \(R\) is the resultant amplitude, match column I with column II.

Given, \(x_1=A_1 \sin \left(\omega t+\phi_1\right)\) and \(x_2=A_2 \sin \left(\omega t+\phi_2\right)\)
A: The two SHMs are in phase, \(A_1=A_2=A\)
\(R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \delta}\) \(=A \sqrt{2(1+\cos \delta)}=\) \(2 A \cos \left(\frac{\delta}{2}\right)\)
where, the phase difference is \(\delta=0\).
\(R=2 A\)
Therefore, A - III.
\(\mathrm{B}\) :The two SHMs are in phase, \(A_1 \neq A_2\) and phase difference \(\delta=0\).
\(R=\sqrt{A_1{ }^2+A_2{ }^2+2 A_1 A_2 \cos \delta} \text { and }\) \(\cos \delta=\cos 0^{\circ}=1 \)
\( \Rightarrow R=\sqrt{A_1^2+A_2^2+2 A_1 A_2}=\left(A_1+A_2\right)\)
Therefore, B - I.
C: The two SHMs are \(90^{\circ}\) out of phase, \(A_1=A_2=A\) and phase difference \(\delta=90^{\circ}\).
\(R=\sqrt{A_1^2+A_2^2+2 A_1 A_2 \cos \delta}\) \(=\sqrt{A^2+A^2}=A \sqrt{2}\)
Therefore, C - IV.
\(\mathrm{D}\) :The two SHMs are \(180^{\circ}\) out of phase, \(A_1=A_2\) and phase difference \(\delta=180^{\circ}\).
\(R=\sqrt{A_1{ }^2+A_2{ }^2+2 A_1 A_2 \cos \delta}=\) \(\sqrt{A^2+A^2-2 A^2}=0\)
Therefore, D - II.