Question

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

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

Answer 1

The correct option is B

${120}^{\circ }$

Step 1: Equations of the S.H.M.

1. Consider the amplitude of the first particle is $A$, the natural frequency is $\omega$, and the time period is $t$.
2. The equation for the S.H.M. for the first particle will be as follows:

$X_1=A\sin \left(\omega t\right)$

3. The amplitude, frequency, and time period of the oscillation are the same for the second particle. Consider the phase difference to be $\varphi$.

4. Hence, the equation for the S.H.M. for the second particle will be as follows:

$X_2=A\sin \left(\omega t+\varphi \right)$

Step 2: The two particles cross each other without collision

The crossing of the particles will happen, when $X_1=\frac{A}{2}$.

$$\begin{gathered} \frac{A}{2}=A\sin \left(\omega t\right) \\ \sin \left(\omega t\right)=\frac{1}{2} \\ \omega t={\sin }^{-1}\left(\frac{1}{2}\right) \\ \omega t={30}^{\circ } \end{gathered}$$

Step 3: Calculation of the phase difference

Since the second particle moves in a different direction,

$$\begin{gathered} \left(\omega t+\varphi \right)=\left({180}^{\circ }-{30}^{\circ }\right) \\ {30}^{\circ }+\varphi ={150}^{\circ } \\ \varphi ={120}^{\circ } \end{gathered}$$

Therefore, the phase difference between the particles is ${120}^{\circ }$.

Hence, option (B) is correct.