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# A block of mass m connected by a spring is kept on a horizontal frictionless surface.When the spring is at its natural length l0,the block is at a distance a from a wall.Now the block is compressed by a distance of 2a  and released.If the collisoin between block and wall is elastic,determine the time period of oscillation

Arun
25763 Points
3 years ago
As we have elongated the spring, such that the maximum distance of block from wall is 2a and equillibrium is at distance a from the wall.

So, the block will not hit the wall when it is released from 2a. It will act as if there is no wall there.
Therefore, the time period of oscillation is given by simply.

T = 2(pi) sqrt (m/k)
Shivang
13 Points
3 years ago
Time period to cover the distance from the mass from the wall after mass is compressed to distance to a from the wall is =T/4 +T/12 =T/3Now,Net force =mz2Ka -ka = mzKa=mz. Here z is accelerationSo,z=ka/mSo we got the relation between z and a So time period to complete 1 oscillation is equal to 2pi/3 (M/k)1 ×2
Rishi Sharma
one year ago
Dear Student,

When spring is stretched, extension in spring x = 2a-a = a.
Here, restoring force balances the SHM force,
maw^2 - kx = 0
w^2 = ka / ma
w = √(k/m)
Time period is given by-
T = 2π/w
T = 2π √(m/k)

Thanks and Regards