Question

Two particles $A$ and $B$ perform SHM along the same straight line with the same amplitude ${}^{\prime }a{}^{\prime },$ same frequency ${}^{\prime }f{}^{\prime }$ and same equilibrium position ${}^{\prime }O{}^{\prime }.$ The greatest distance between them is found to be $\frac{3a}{2}$. At some instant of time they have the same displacement from mean position. What is the displacement?

• A. $\frac{a}{2}$
• B. $\frac{a\sqrt{7}}{4}$
• C. $\frac{\sqrt{3}}{a2}$
• D. $\frac{3a}{4}$

Answer 1

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

The two particles will have maximum distance = $3a/2$
when one particle is at the extreme position and other is at $a/2$ distance.
The phase difference between the two particles is ${120}^{\circ }$.

Since the particles have equal frequency the phase angle travelled by two particles will be equal in equal interval of time.

here, $y_1=y_2$

or $a\sin (\omega t+\theta +120)=a\sin (\omega t+\theta )\Rightarrow \omega t+\theta =30$${}^{\circ }$

Thus, $y_1=a\sin \left(30\right)=\frac{a}{2}$