question_answer

Two particles execute simple harmonic motions of the same amplitude and frequency along the same straight line. They cross one another when going in opposite directions. What is the phase difference between them when their displacements are half of their amplitudes?

Answer:

                The general equation for S.H.M. is \[y=A\sin (\omega t+{{\phi }_{0}})\] As the displacement is half of the amplitude \[(y=A/2),\]so \[A/2=A\sin (\omega t+{{\phi }_{0}})\] or  \[\sin (\omega t+{{\phi }_{0}})=\frac{1}{2}\] \[\therefore \]  \[\omega t+{{\phi }_{0}}={{30}^{\circ }}\] or \[{{150}^{\circ }}\] As the two particles are going in opposite directions, the phase of one is \[\text{3}0{}^\circ \] and that of the other \[\text{15}0{}^\circ \] Hence the phase difference between the two particles \[=150-30=\mathbf{12}{{\mathbf{0}}^{\mathbf{o}}}\]