Figure shown a block P of mass m resting on a smooth horizontal surface, attached to a spring of force constant k which is rigidly fixed on the wall on left side, shown in the figure . At a distance l to the right of the block there is a rigid wall. If block is pushed towards loft so that spring is compressed by a distance `5l//3` and when released, if will starts its oscillations. If collision of block with the wall is considered to be perfectly elastic. Find the time period of oscillation of the block.

Explanation

As shown in figure , as the block its released from rest at a distance `5//3` from its mean position, this will be the amplitude of oscillation. But on the other side of mean position block can move only up to a distance `l` from mean position and then it will return from this point with equal velocity due to elastic collision . Consider figure . If no right wall is present during oscillation, block P will excute complete SHM on right side of mean position also up t its amplitude `5l//3` . Thus, we can observe the block P at a point X at distance l from mean position (where in our case wall is present) . if block passes this position at speed v (which is the speed with , which block bites the wall in our case) after reaching its exterme position y, it willl return and during return path it will cross the position X with the same speed v (as displacement from mean position is same).

Thus, we can state that in our case of given problem , that block P is executing SHm but it skips a XYX on right half of its motion due to electric collision of the block with the wall. if `T_(0)` is the time period of this oscillation we look at figure which shown the corresponding circular motion repersantation. Here during oscillation of point P partical `P_(0)` covers its circular motion along `ECDAF`, and from F it instantly jumps to E (due to elastic collision of P with wall at X) and again carry out ECDAF and so on thus we can find the time of this total motion as

`t = (pi + 2 theta)/(omega) = (pi + 2 sin ^(-1) (3//5))/(sqrt((k)/(m))`
`= sqrt((m)/(k)) [pi + 2 sin ^(-1) ((3)/(2))]`
We can also find the time period of this motion by sub-tracting the time FYE from the total time period as
`t =(2 pi - 2 cos^-1 ((3)/(5)))/(omega)`
`= sqrt((m)/(k)) [pi + 2 cos^(1) ((3)/(5))]`
Equations (i) and (ii) will result same numerical value.