A body `A` of mass `m_(1)=1kg` and a body `B` of mass `m_(2)=4kg` are attached to the ends of a spring. The body `A` performs vertical simple harmonic oscillations of amplitude `a=1.6 cm` and angular frequency`omega=25rad/s`. Neglecting the mass of the spring determine the maximum and minimum values of force the system exerts on the surface on which in rests.
`["Take g"=10 m//s^(2)]`

Explanation

While the body `A` is at its upper extreme position, the spring is obviously elongated by the amount
`(a-(m_(1)g)/(k))`
If we indicate `y-` axis in vertically downward direction, Newton's second law of motion in projection form i.e.`F_(y)=m w _(y)` for body `A` gives :
`m_(1)g+k(a-(m_(1)g)/(k))=m_(1)omega^(2)a` or, `k(a-(m_(1)g)/(k))=m_(1)(omega^(2)a-g) (1)`
(Because at any extreme position the magnitude of acceleration of an oscillating body equals `omega^(2)a` and is restoring in nature.)
If `N` be teh normal force exerted by the floor on the body `B`, while the body `A` is at its upper extreme position, from Newton's second lw for body `B`
`N+k(a-(m_(1)g)/(k))=m_(2)g`
or, `N=m_(2)g-k(a-(m_(1)g)/(k))=m_(2)g-m_(1)(omega^(2)a - g)` ( using Eqn.1)
Hence `N=(m_(1)+m_(2))g-m_(1)omega^(2)a`
When the body `A` is at its lower extreme position, the spring is compressed by the distance `(a+(m_(1)g)/(k))`.
From Newton's second law in projection from i.e.`F_(y)=m w_(y)` for body `A` at this state `:`
`m_(1)g=k(a+(m_(1)g)/(k))=m_(1)(-omega^(2)a)` or, `k(a+(m_(1)g)/(k))=m_(1)(g+omega^(2)a) .....(3)`
In this case if `N^(')` be the normal force exerted by the floor on the body `B`, From Newton's second law
for body `B` we get : `N^(')=k(a+(m_(1)g)/(k))+m_(2)g=m_(1)(g+omega^(2)a)+m_(2)g` (using En. 3)
Hence `N^(')=(m_(1)+m_(2))g+m_(1)omega^(2)a`
From Newton's third law the magnitude of sought forces are `N^(')` and `N`, respectively.