Time period `(T)` and amplitude `(A)` are same for two particle which undergoes `SHM` along the same line. At one particular instant one particle is at phase`(3pi)/(2)`and the other is at zero, while moving in the same direction. Find the time at which they will cross each other

Options

A. `4T//2`
B. `3T//8`
C. `3T//4`
D. `3T//7`

Answer

B

Explanation

At phase `3pi//2, (omegat + phi) = 3pi//2`
`x_(1) = A sin'(3pi)/(2) = -A`
`x_(2) = 0`
So at `t = 0 , x_(1) = -A cos omega` and `x_(2) = A sin omega t`
` x_(1) =x_(2)`
`-cos omegat = sin omegat`
`tan omegat = -1`
`omega t = (2pi)/(4)`
`((2pi)/(T)) t = (3pi)/(4)`
`t = 3T//8`