A small block of mass m is kept on a bigger block of mass M which is attached to a vertical spring of spring constant k as shown in the figure. The system oscilates verticaly. a.Find the resultant force on the smaller block when it is displaced through a distance x above its equilibrium position. b. find the normal force on the smaller blok at this position. When is this force smallest smaller block at this position. When is this force smallest in magnitude? c. What can be the maximum amplitude with which the two blocks may oscillate together?

Answer

`x=(mkx)/(M+m)``=
-mg-(mkx)/(M+N)`
maximum amplitude
`=(g(M+m))/k`

Explanation

`a`. From the fre bdoy diagram
:. `R+momegay^2x-mg=0` ……….
Resultant force `momega^2x=mg-R`
`rarr momega^2x=m(k/(M+m))x`
`x=(mkx)/(M+m)`
`omega=sqrt({k/(M+m)})`
[for spring mas system]
`b`. `R=mg-momega^2x`
`=mg-m(mk)/(M+N)x`
`=-mg-(mkx)/(M+N)`
For LR to be smallest `momega^2x` should be max, i.e. x is maximum.
The particle should be at the highest point.
`c`. We have `R=mg-momega^2x`
The two blocks may oscillate together in such a way that R is greater than 0.
At limiting condition R=0
`mg=momega^2x`
`=x=(mg)/(momega^2)=(mg.(M+m))/(mk)`
So the maximum amplitude
`=(g(M+m))/k`