Question

Two particle execute S H M of same amplitude and frequency along the same straight line, They pass one another going in opposite direction each time their displacement is half of their amplitude. 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 ${120}^{\circ }$

$\begin{array}{c}Accordingtoquestion.............\\ Infirstcase:\\ \frac{a}{2}=asin\omega t\\ \Rightarrow sin\omega t=\frac{1}{2}\\ andinsecondcase:\\ \frac{a}{2}=asin(\omega t+\phi )\\ \Rightarrow sin(\omega t+\phi )=\frac{1}{2}\\ Now,\\ \Rightarrow sin\omega t\cos \phi +\cos \omega t\sin \varphi =\frac{1}{2}\\ \Rightarrow \frac{1}{2}\cos \phi +\sqrt{1-\frac{1}{4}}\sin \phi =\frac{1}{2}\\ \Rightarrow \sqrt{3}\sin \varphi =1-\cos \phi \\ \Rightarrow {\sin }^2\phi =1+{\cos }^2\phi -2\cos \phi \\ \Rightarrow 4{\cos }^2\phi -2\cos \phi -2=0\\ \Rightarrow \cos \phi =\frac{1-3}{4}=-\frac{1}{2}\\ \therefore \cos \phi =-\frac{1}{2}\\ \phi ={120}^0\\ sothatthecorrectoptionisB.\end{array}$