Question

Two particles execute 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.
• B.
• C.
• D.

Answer 1

The correct option is D

Let two simple harmonic motions are $y=Asin\omega tand{y}^{\prime }=Asin(\omega t+\phi )$
In the first case $\frac{A}{2}=Asin\omega t\Rightarrow sin\omega t=\frac{1}{2}$ $\therefore cos\omega t=\frac{\sqrt{3}}{2}$
In the second case $\frac{A}{2}=Asin(\omega t+\phi )$
$\Rightarrow \frac{1}{2}=[sin\omega t.cos\phi +cos\omega tsin\phi ]\Rightarrow \frac{1}{2}=[\frac{1}{2}cos\phi +\frac{\sqrt{3}}{2}sin\phi ]$
$\Rightarrow 1-cos\phi =\sqrt{3}sin\phi \Rightarrow (1-cos\phi {)}^2=3si{n}^2\phi \Rightarrow (1-cos\phi {)}^2=3(1-co{s}^2\phi )$
By solving, we get $cos\phi =+1orcos\phi =-\frac{1}{2}$
i.e. $\phi =0$ Or $\phi ={120}^{\circ }$