Figure shown a spring block system hanging in equilibrium. The block of system is pulled down by the distance x and imparted a velocity v in downward direction as shown in figure. Find the time it will take to reach its mean position. Also find the maximum distance to which if will move before returning back towards mean position.

Explanation

As shown in figure (b), when the block is pulled down by a distance x and throwndownward , it will start executting SHM. It will go further to a distance A (amplitude) from mean possion before returning back v at a displacement x from its mean position as
`v = omega sqrt(A^(2) - x^(2))`
or `v^(2) = (k)/(m) (A^(2) - x^(2))`
[as for spring block system `omega = sqrt((k)/(m)) ` ltbr or `A = sqrt((mv^(2))/(k) + x^(2)))`
Now timeof motion of bob can be obtained by circular motion representation of the respectiveSHM is shown in figure at `t = 0 At t = 0`, block P is at a distance x force its mean position in downward direction and it is moving downwards so we can sider its corresponding circular motion partical in III quadrant as shown in reference angular velocity . We consider P will reach its mean position when partical `P_(0)` reaches position A by traversing an angle `theta` as shown in the figure.

Thus it will take a time gives as `t = (theta)/(omega) = (pi - sin^(-1) ((x//A)/(sqrt((k)/(m))))`
or `t = sqrt((m)/(k)) [pi - sin^(-1) ((x)/(sqrt((mv^(2))/(k) + x^(2))))]`
The maximum distance to which block will move from its initial position is `A - x` as it gas upto its lower exterme at a distance equal to its amplitude A after to mean position.