Question

Two particles are in S.H.M with same amplitude 'a', same frequency 'f' and along the same straight line with the same mean position. The maximum distance between the two particles is $a\sqrt{2}$. If the phase difference between the two particles is $\frac{\pi }{x}$rad, then x is

• A. $2$
• B. $4$
• C. $1$
• D. $5$

Answer 1

The correct option is A $2$

Let the SHM be along the vertical direction and the center of the circle is the mean position. $2\theta$ represents the phase difference between the particles which is $\frac{\pi }{x}$.

The maximum distance between the two particles would be when both the particles are at equal distances from the mean position in opposite direction. We have $A{A}^{\prime }=Q{Q}^{\prime }=a\sqrt{2}$

So, we get $sin\theta =\frac{a}{a\sqrt{2}}=\frac{1}{\sqrt{2}}$. (The radius of the circle is equal to the amplitude $a$)

Thus we get $\theta =\frac{\pi }{4}$ and thus the phase difference is $\frac{\pi }{2}$ giving $x=2$