Question

Two particles execute $SHM$ of the same amplitude and frequency, along the same straight line and from the same mean position. They cross one another without collision when going in opposite directions each time the displacement from mean position is half of their amplitudes. The phase-difference between them is,

• A. $0^{\circ }$
• B. ${120}^{\circ }$
• C. ${180}^{\circ }$
• D. ${135}^{\circ }$

Answer 1

The correct option is B ${120}^{\circ }$

Let the equations of motion of the two particles are,
$x_1=a\sin (\omega t+{\varphi }_1)$
And, $x_2=a\sin (\omega t+{\varphi }_2)$

Both the particles cross each other at $x=\frac{a}{2}$

Let particle $\left(1\right)$ is travelling towards positive extremity.
For particle $\left(1\right)$, $\frac{a}{2}=a\sin (\omega t+{\varphi }_1)$

After solving, we get, $(\omega t+{\varphi }_1)={30}^{\circ }$

Particle $\left(2\right)$ is travelling towards negative extremity.
For particle $\left(2\right)$, $\frac{a}{2}=a\sin (\omega t+{\varphi }_2)$

After solving, we get, $(\omega t+{\varphi }_2)={150}^{\circ }$

As particle $\left(2\right)$ is travelling in opposite direction, its phase must be greater than ${90}^{\circ }$.

$\therefore$ phase difference, $\Delta \varphi ={150}^{\circ }-{30}^{\circ }={120}^{\circ }$

Hence, $\left(B\right)$ is the correct answer.