Question

Two particles executes S.H.M. 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. 30⁰
• B. 60⁰
• C. 90⁰
• D. 120⁰

Answer 1

Answer: (D)

120⁰

Answer 2

Let two simple harmonic motions are $y = a\sin\omega t$ and $y = a\sin(\omega t + \varphi)$

In the first case $\frac{a}{2} = a\sin\omega t$ ⇒ $\sin\omega t = 1/2$

∴ $\cos\omega t = \frac{\sqrt{3}}{2}$

In the second case $\frac{a}{2} = a\sin(\omega t + \varphi)$

⇒ $\frac{1}{2} = \lbrack\sin\omega t.\cos\varphi + \cos\omega t\sin\varphi\rbrack$

⇒ $\frac{1}{2} = \left\lbrack \frac{1}{2}\cos\varphi + \frac{\sqrt{3}}{2}\sin\varphi \right\rbrack$

⇒ $1 - \cos\varphi = \sqrt{3}\sin\varphi$⇒ $(1 - \cos\varphi)^{2} = 3\sin^{2}\varphi$

⇒$(1 - \cos\varphi)^{2} = 3(1 - \cos^{2}\varphi)$ By solving we get $\cos\varphi = + 1$ or $\cos\varphi = - 1/2$

i.e. $\varphi = 0$ or $\varphi = 120^{o}$