Question

Two particles ‘A’ and ‘B’ start SHM at t = 0. Their positions as a function of time are given by
$X_A=A\sin \omega t$
$X_B=A\sin (\omega t+\frac{\pi }{3})$
$\begin{array}{cc}\text{Column - I}& \text{Column - II}\\ \text{(A) Minimum time when x is same}& \left(P\right)\frac{5\pi }{6\omega }\\ \text{(B) Minimum time when velocity is same}& \left(Q\right)\frac{\pi }{3\omega }\\ \text{(C) Minimum time after which}{v}_A<0\text{and}{v}_B<0& \left(R\right)\frac{\pi }{\omega }\\ \text{(D) Minimum time after which}{x}_A<0\text{and}{x}_B<0& \left(S\right)\frac{\pi }{2\omega }\end{array}$
Which of the following option has the correct combination considering column-I and column-II.

• A. $B\to S$
• B. $C\to Q$
• C. $D\to R$
• D. $A\to P$

Answer 1

The correct option is C $D\to R$

(A)-Q, (B)-P, (C)-S, (D)-R