A block of mass m is pushed against a spring of spring constant k fixed at ne end to a wall. The block ocan side on a frictionless tableas shown in figure. The natural length of thespring is `L_0` and it is compressed ti half its natural length when the block is relesed. Find teh velocity of the block aa s function of its distance x from the wall .

Explanation

Figure the question, we find that the spring pushes the block to the right till the spring acquires its natural length `(L_(0))`. At this stage, the block loses contact with the spring and moves with a constant velocity `v^(')` on the frictionless table.
Now, inital compression of the spring `=(L_(0))/(2)`.
When block is at a distance x from the wall and `x lt L_(0)`, the compression is `(L_(0)-x)` and velocity of block is v
Applying the principle of conservation of energy, we get
`(1)/(2)k(L_(0)/(2))^(2)=(1)/(2)k(L_(0)-x)^(2)+(1)/(2)mv^(2)`
or `(1)/(2)mv^(2)=(1)/(2)k[(L_(0)/(2))^(2)-(L_(0)-x)^(2)]`
`v=sqrt((k)/(m))[(L_(0)^(2))/(4)-(L_(0)-x)^(2)]^(1//2)`