Question

Two particles A and B perform SHM along the same straight line with the same amplitude 'a', same frequency 'f' and same equilibrium position 'O'. The greatest distance between them is found to be 3a/2. at some instant of time they have the same displacement from mean position. What is this displacement?

• A. $\frac{a}{\sqrt{2}}$
• B. $a\sqrt{7}/4$
• C. $\sqrt{7}a/4$
• D. 3a/4

Answer 1

The correct option is A $\frac{a}{\sqrt{2}}$

Since the largest distance between them is $\frac{30}{2}$, hence the phase difference$={90}^{\circ }+{45}^{\circ }={135}^{\circ }$

$y_1=a\sin (\omega t+{135}^{\circ })y_2=a\sin \left(\omega t\right)y_1=y_2\sin (\omega t+{135}^{\circ })=\sin \left(\omega t\right)$

$\therefore \omega t=\pi t=\frac{\pi }{\omega }{y}_1=a\sin (\omega t+{135}^{\circ })=a\sin (\omega \times \frac{\pi }{\omega }+{135}^{\circ })=-\frac{a}{\sqrt{2}}$