Question

Two particles are in SHM with same amplitude $A$ and same angular frequency $\omega$. At time $t=0$, one is at $x=+\frac{A}{2}$ and other is at $x=-\frac{A}{2}$. Both are moving in same direction.

• A. Phase difference between the two particle is $\frac{\pi }{3}$
• B. Phase difference between the two particle is $\frac{2\pi }{3}$
• C. They will collide after time $t=\frac{\pi }{2\omega }$
• D. They will collide after time $t=\frac{3\pi }{\omega }$

Answer 1

The correct option is A Phase difference between the two particle is $\frac{\pi }{3}$

Let $x_1$ be the position of first particle and $x_2$ be of the second particle:

$x_1=A\sin (\omega t+{\varphi }_1)$

$x_2=A\sin (\omega t+{\varphi }_2)$

At $t=0$, for first particle:

$x_1=\frac{A}{2}=A\sin \left({\varphi }_1\right)$

$\Rightarrow \sin \left({\varphi }_1\right)=\frac{1}{2}$

$\Rightarrow {\varphi }_1=\frac{\pi }{6}$, or ${\varphi }_1=\frac{5\pi }{6}$

Similarly for the second particle,

$x_2=-\frac{A}{2}=A\sin \left({\varphi }_2\right)$

$\Rightarrow \sin \left({\varphi }_2\right)=-\frac{1}{2}$

$\Rightarrow {\varphi }_2=-\frac{\pi }{6}$, or ${\varphi }_2=-\frac{5\pi }{6}$

Now for both of the particles to be moving in the same direction, at $t=0$:

$v_1=A\omega \cos \left({\varphi }_1\right)$

$v_2=A\omega \cos \left({\varphi }_2\right)$

Case I: ${\varphi }_1=\frac{\pi }{6}$, and ${\varphi }_2=-\frac{\pi }{6}$, where particle 1 leads particle 2.

Then,

$\Delta \varphi ={\varphi }_1-{\varphi }_2=\frac{\pi }{3}$

Case II: ${\varphi }_1=\frac{5\pi }{6}$, and ${\varphi }_2=-\frac{5\pi }{6}$, where particle 2 leads particle 1.

Then,

$\Delta \varphi ={\varphi }_2-{\varphi }_1=-\frac{10\pi }{6}=-\frac{5\pi }{3}=\frac{\pi }{3}$

Now, to collide, the combined angular displacement of both particles must cover-up this phase lag, that is in time $t$:

$\omega t+\omega t=\Delta \varphi$

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

$\Rightarrow t=\frac{\pi }{6\omega }$