Q162 / 200 · Manipal (MET) 2014 · Physics
Two particles executes SHM of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions. Each time their displacement is half of their amplitude. The phase difference between them is
- A$30^{\circ}$
- B$60^{\circ}$
- C$90^{\circ}$
- Correct answerD$120^{\circ}$
Solution
Let two simple harmonic motions are $y=a \sin \omega t$ and $y=a \sin (\omega t+\phi)$
In the first case, $\frac{a}{2}=a \sin \omega t \Rightarrow \sin \omega t=\frac{1}{2}$
$\therefore \quad \cos \omega t=\frac{\sqrt{3}}{2}$
In the second case, $\frac{a}{2}=a \sin (\omega t+\phi)$ $\begin{aligned} & \Rightarrow \quad \frac{1}{2}=[\sin \omega t \cdot \cos \phi+\cos \omega t \sin \phi] \\ & \Rightarrow \frac{1}{2}=\left[\frac{1}{2} \cos \phi+\frac{\sqrt{3}}{2} \sin \phi\right] \\ & \Rightarrow \quad 1-\cos \phi=\sqrt{3} \sin \phi \\ & \Rightarrow \quad(1-\cos \phi)^2=3 \sin ^2 \phi \\ & \Rightarrow \quad(1-\cos \phi)^2=3\left(1-\cos ^2 \phi\right) \end{aligned}$
On solving we get $\begin{aligned} & \cos \phi=+1 \text { or } \cos \phi=\frac{-1}{2} \\ & \text { i.e., } \quad \phi=0 \text { or } \phi=120^{\circ} \end{aligned}$
In the second case, $\frac{a}{2}=a \sin (\omega t+\phi)$ $\begin{aligned} & \Rightarrow \quad \frac{1}{2}=[\sin \omega t \cdot \cos \phi+\cos \omega t \sin \phi] \\ & \Rightarrow \frac{1}{2}=\left[\frac{1}{2} \cos \phi+\frac{\sqrt{3}}{2} \sin \phi\right] \\ & \Rightarrow \quad 1-\cos \phi=\sqrt{3} \sin \phi \\ & \Rightarrow \quad(1-\cos \phi)^2=3 \sin ^2 \phi \\ & \Rightarrow \quad(1-\cos \phi)^2=3\left(1-\cos ^2 \phi\right) \end{aligned}$
On solving we get $\begin{aligned} & \cos \phi=+1 \text { or } \cos \phi=\frac{-1}{2} \\ & \text { i.e., } \quad \phi=0 \text { or } \phi=120^{\circ} \end{aligned}$
Hence, the correct answer is option: 'D'.