Two particles execute SHM of same time period but different amplitudes along the same line. One starts from mean position having amplitude A and other starts from extreme position having amplitude 2A. Find out the time when they both will meet?

Explanation

We solve the above problem with the help of phasor diagram. First we draw the initial position of both the particle on the phasor. From figure phase difference between both the particle is `pi`/2.
They will meet each other when their projection from the circle on the horizontal diameter meet at one point.
Now from figure :
EF = A cos `theta = 2A sin theta `
tan `theta = (1)/(2)`
`theta^(-1) ((1)/(2))`
So time taken by the particle to cross each other
`t = ("angle travelled by A")/(omega ) implies t = (pi//2-theta)/(omega)`