In the circuit shown in the figure, if potentail at point `A` is taken to be zero, the potential at point `B` is

Answer

`x_(2) = v_(0)t + (m_(1))/(m_(2)) A (1 - cos omegat), l_(0) = ((m_(1))/(m_(2)) + 1) A`

Explanation

The string snaps and the spring force comes into play. The spring force beings an internal force for the two mass-spring system will not be able to change the velocity of centre of mass. This means the location of centre of mass at time t will be `v_(0)t`
`x_(cm) = (m_(1)x_(1) + m_(2)x_(2))/(m_(1) + m_(2)) = v_(0)t`
`rArr m_(1)[v_(0)t - A(1 - cos omegat)] + m_(2)x_(2) = v_(0)tm_(1) + v_(0)tm_(2)`
`rArr m_(2)x_(2) = v_(0)tm_(1) + v_(0)tm_(2) - v_(0)tm_(1) + m_(1)A (1 - cos omegat)`
`rArr m_(2)x_(2) = v_(0)tm_(2) + m_(1)A(1 - cos omegat)`
`rArr x_(2) = v_(0)t + (m_(1))/(m_(2)) A(1 - cos omegat)`
(b) Given that `x_(1) = v_(0)t - A(1 - cos omegat)`
`:. (dx_(1))/(dt) = v_(0) - Aomega sin omegat`
`:. (d^(2)x_(1))/(dt^(2)) = -Aomega^(2) cos omegat` ....(i)
This is the acceleration of mass `m_(1)`. When the spring comes to its natural length instantaneously then `(d^(2)x_(1))/(dt^(2)) = 0` and `x_(2) - x_(1) = l_(0)`
`:. [v_(0)t + (m_(1))/(m_(2)) A(1 - cos omegat)] - [v_(0)t - A (1 - cos omegat)] = l_(0)`
`((m_(1))/(m_(2)) + ) A ( 1 - cos omega t ) = l_(0)`
Also when `(d^(2)x_(1))/(dt^(2)) = 0, cos omegat = 0` from (1)
`:. l_(0) = ((m_(1))/(m_(2)) + 1) A`