Question

Two SHM particles `P_(1)` and `p_(2)` start from `+ (A)/(2)` and `-sqrt(3A)/(2)`, both in negative directions. Find the time (in terms of T) when they collide. Both particles have same omega, `A` and `T` and the execute SHM along the same line.

Answer 1

Correct Answer - A
In the first figure, we have shown two particles `Q_(1)` and `Q_(2)` corresponding to given initial conditions of `P_(1)` and `P_(2)`. In the second figure, we have shown the moment when the SHM particles are colliding at `x = -A cos 45^(@) = -(A)/(sqrt2)`.
We can see that either of the particles `Q_(1)` and `Q_(2)` has rotated an angle `theta = 75^(@)` or `(5pi)/(12)`. So, the time taken is
`t = (theta) /(omega) = ((5pi//12))/((2pi//T)) = ((5)/(24))T`
.