Two particles execute simple harmonic motion along the same straight line with same amplitude and same…
Two particles execute simple harmonic motion along the same straight line with same amplitude and same frequency. The two particles pass one another when moving in opposite directions each time at a distance of $\frac{1}{\sqrt{2}}$ times the amplitude from their common mean position. The phase difference between the two particles is
$30^{\circ}$
$45^{\circ}$
$60^{\circ}$
$90^{\circ}$
Solution
Position of particle $1, x_1=A \sin \phi_1$
Position of particle 2, $x_2=A \sin \phi_2$
let particle 1 is moving away from mean position
Given, $x_1=x_2=\frac{A}{\sqrt{2}}$
$\therefore \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \sin \phi_1 \Rightarrow \phi_1=\frac{\pi}{4}$
let particle 2 is moving towards mean position,
$\therefore \frac{\mathrm{A}}{\sqrt{2}}=\mathrm{A} \sin \phi_2 \Rightarrow \phi_2=\frac{3 \pi}{4}$
Phase difference, $\left|\phi_1-\phi_2\right|=\left(\frac{\pi}{4}-\frac{3 \pi}{4}\right)=\frac{\pi}{2}$
$\phi=90^{\circ}$