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 the displacement?

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

Answer 1

The correct option is B $a\sqrt{7}/4$

Let the initial phase difference be $\theta$, so diference in displacement is $asin\omega t-asin(\omega t+\theta )=2asin\left(\theta /2\right)cos(\omega t+\theta /2)$

Maximum displacement is $2a\left(sin\right(\theta /2\left)\right)=3a/2\Rightarrow \theta =2si{n}^{-1}\left(3/4\right)$

When both have same displacement from mean position, $\omega t+\theta =\pi -\omega t\Rightarrow \omega t=(\pi -\theta )/2$

So this displacement is $asin(\pi /2-si{n}^{-1}(3/4\left)\right)=acos\left(si{n}^{-1}\right(3/4\left)\right)=a\frac{\sqrt{7}}{4}$