Question

Two particles are in SHM in a straight line about same equilibrium position.Amplitude $A$ and time period $T$ of both the particles are equal.At time $t=0,$ one particle is at displacement $y_1=+A$ and the other at $y_2=\frac{A}{2}$, and they are approaching towards each other. After what time they cross each other?

• A. $\frac{T}{3}$
• B. $\frac{T}{4}$
• C. $\frac{5T}{6}$
• D. $\frac{T}{12}$

Answer 1

The correct option is D $\frac{T}{12}$

$y_1\left(t\right)=Acos\left(\omega t\right)$

$y_2\left(t\right)=Acos(\omega t+\theta )$

At $y_2\left(0\right)=A/2$

$⟹\theta =-\pi /3$ or $\pi /3$

Now,

$v_1\left(0^{+}\right)=-A\omega sin\left(\omega \right(0^{+}\left)\right)<0$

$v_2\left(0\right)=-A\omega \left(sin\right(\theta \left)\right)$

Given $v_1\left(0\right)v_2\left(0\right)<0$

$⟹v_2\left(0\right)>0i.e.\theta =-\pi /3$

$y_1\left(t\right)=y_2\left(t\right)$

Solving, $tan\left(\omega t\right)=1/\sqrt{3}$

$⟹t=T/12$