A particle of mass `m` attached with a massless spring natural length `l` and stiffness `k` is released from rest from the horizontal position of the relaxed spring. When the particle passes through its lowest point, the maximum length of the spring becomes `2l`. Find the:

a. speed of the particle at its lowest point.
b. acceleration of the particle at its lowest position.

Answer

a. sqrt(4gl-K/ml^2)`; b. `(Kl-mg)/(m)`

Explanation

At B, extension in the spring=l
a. `mg*2l=1/2Kl^2+1/2mv^2`
`impliesv=sqrt(4gl-K/ml^2)`

b. `Kl-mg=maimpliesa=(Kl-mg)/(m)`