A smooth rubber cord of length `l` whose coefficient of elasticity is k is suspended by one end from the point O (figure). The other end is fittetd with a catch B. A small sleeve A of mass m starts falling from the point O. Neglecting the masses of the thread and the catch, find the maximum elongation of the cord.

Explanation

Suppose that `Deltal` is the elongation of the rubber cord. Then from energy conservation,
`DeltaU_(gr)+DeltaU_(el)=0` (as `DeltaT=0`)
or, `-mg(l+Deltal)+1/2kDeltal^2=0`
or, `1/2kDeltal^2-mgDeltal-mgl=0`
or, `Deltal=(mg+-sqrt((mg)^2+4xxk/2mgl))/(2xxk/2)xxk/2=(mg)/(k)[1+sqrt(1+-(2kl)/(mg))]`
Since the value of `sqrt(1+(2kl)/(mg))` is certainly greater than 1, hence negative sign is avoided.
So, `Deltal=(mg)/(k)(1+sqrt(1+(2kl)/(mg)))`