Question

Two particles A and B execute SHM along the same line with the same amplitude a,same frequency and same equilibrium position O. If the phase difference between them is $\theta =2{\sin }^{-1}\left(0.9\right)$,then find the maximum distance between the two

• A. $1.8a$
• B. $0.9a$
• C. $2.7a$
• D. $1.5a$

Answer 1

The correct option is A $1.8a$

Let the phase difference be $\theta$

$y_1=a\sin \left(\omega t\right)$

$y_2=a\sin (\omega t-\theta )$

$y_1-y_2=a\sin \omega t-a\sin (\omega t-\theta )=2a\left(\sin \right(\theta )\times \cos (2\omega t-\varphi \left)\right)$

max$(y_1-y_2)=1.8a$ because max of $cos(2\omega t-\theta )=1$