Question

Two particles move parallel to x-axis about the origin with same amplitude ‘a’ and frequency $\omega$. At a certain instant they are found at a distance a/3 from the origin on opposite sides but their velocities are in the same direction. The phase difference between the two particles is..

• A. $co{s}^{-1}\frac{7}{9}$
• B. $co{s}^{-1}\frac{5}{9}$
• C. $co{s}^{-1}\frac{4}{9}$
• D. $co{s}^{-1}\frac{1}{9}$

Answer 1

The correct option is A $co{s}^{-1}\frac{7}{9}$

Let$x_1=asin\omega tand{x}_2=asin(\omega t+\delta )betwoS.H.M\frac{a}{3}=asin\omega tand\frac{-a}{3}=asin(\omega t+\delta )sin\omega t=\frac{1}{3}andsin(\omega t+\delta )=\frac{-1}{3}Eliminatingt,\frac{1}{3}\text{cos}\delta +\sqrt{1-\frac{1}{9}}\text{sin}\delta =\frac{-1}{3}9co{s}^2\delta +2cos\delta -7=0cos\delta =-1or\frac{7}{8}i.e.,\delta ={180}^{\circ }orco{s}^{-1}\left(\frac{7}{9}\right)Now{v}_1=a\omega cos\omega tand{v}_2=a\omega cos(\omega t+\delta )Ifweput\delta ={180}^{\circ }Wefindthat{v}_{1}and{v}_{2}areofoppositesigns.Hence\delta ={180}^{\circ }isnotapplicable.\therefore \delta =co{s}^{-1}\frac{7}{9}$