Question

Two particles are executing SHM in a straight line with same amplitude $A$ and time period $T$. At time $t=0$, one particle is at displacement $x_1=+A$ and the other at $x_2=-A/2$ and they are approaching towards each other. After what time they cross each other?

• A. $\frac{T}{3}$
• B. $\frac{T}{6}$
• C. $\frac{T}{9}$
• D. $\frac{T}{12}$

Answer 1

Answer: (B)

$\frac{T}{6}$

We know that,

For SHM

$x=Asin(\omega t+\varphi )...\left(i\right)$

For particle 1

at t= 0, x = +A

putting on eqn (i)

$A=Asin\varphi$

$\Rightarrow sin\varphi =1$

$\Rightarrow \varphi =\frac{\pi }{2}$

hence,

$x=Asin(\omega t+\frac{\pi }{2})...\left(ii\right)$

For particle 2

at $t=0,x=\frac{-A}{2}$

putting on eqn (i)

$\frac{-A}{2}=Asin\varphi$

$\Rightarrow sin\varphi =\frac{-1}{2}$

$\varphi =\frac{-\pi }{6}.$

$v=\frac{dx}{dt}=A\omega cos(\omega t+\varphi )$

at $t=0,$ velocity in of particle 2

$v=a\omega \omega s\left(\frac{-\pi }{6}\right)$

$v=\frac{\sqrt{3}}{2}A\omega >0$

hence, $\varphi =\frac{-\pi }{6}$

so, For particle 2.

$x=Asin(cot-\frac{\pi }{6})...\left(iii\right)$

when both the particle will meet, there

position 'k' will be some

$\Rightarrow Asin(\omega t+\frac{\pi }{2})=Asin(wt-\frac{\pi }{6})$

$\Rightarrow sin(\omega t+\frac{\pi }{2})=sin(\omega t-\frac{\pi }{6})$

$cos\omega t=sin\omega tcos\frac{\pi }{6}-cos\omega tsin\frac{\pi }{6}$

$\Rightarrow cos\omega t=\frac{\sqrt{3}}{2}sin\omega t-\frac{1}{2}cos\omega t$

$\Rightarrow \sqrt{3}sinwt=3cos\omega t\Rightarrow tan\omega t=\frac{3}{\sqrt{3}}=\sqrt{3}$

$\Rightarrow \omega t=\frac{\pi }{3}\Rightarrow \frac{2\pi }{T}\times t=\frac{\pi }{3}\Rightarrow t=\frac{T}{6}$