A plank with a body of mass `m` places on it starts moving straight up according to the law `y=a(1-cos omegat t )`, whee `y` is the displacement from the initial position, `omega =11 s^(-1)`. Find :
(a) the time dependence of the force that the body exerts on the plank if `a=4.0cm,` plot this dependence,
`(b)` the minimum amplitude of oscillation of the plank atwhich the body starts falling behind the plank,
(c) the amplitude of oscillation of the plank at which the body springs up to a height `h=50cm` relative to the initial position (at the moment `t=0)`.

Explanation

For the block from Newton's second law in projection form `F_(y)=m w_(y)`
`N-mg=m ddot(y) ….(1)`
But from `y=a(1-cos omegat)`
We get `ddot(y)=omega^(2)a cos omegat .....(2)`
From Eqns `(1)` and `(2)`
`N=mg(1+(omega^(2)a)/(g)cos omegat) ...(3)`
From Newton's third law the force by which the body `m` exerts on the block is directed vertically downward and equals `N=mg(1+(omega^(2)a)/(g)cos omegat)`
`(b)` When the body `m` starts, falling behind the plank or loosing contact, `N=0`, (because the normal readion is the contact force ). Thus from Eqn. `(3)`
`mg(1+(omega^(2)a)/(g)cos omegat)=0` fro some `t.`
Hence `a_(mi n)=g//omega^(2)~=8 cm`.
(c) We observe that the motion takes place about the mean position `y=a`. At the initial inistant `y=0`. As shown in `(b)` the normal reaction vanishes at a height `(g//omega^(2))` above the position of equilibrium and the body flies off as a free body. The speed of the body at a distance `(g//omega^(2))` from the equilibrium position is `omega sqrt(a^(2)-(g//omega^(2))^(2))`, so that the condition of the problem gives
`([omegasqrt(a^(2)-(g//omega^(2))^(2))])/(2g)+(g)/(omega^(2))+a=h`
Hence solving the resulting quadratic equation and taking the positive roof ,
`a=-(g)/(omega^(2))+sqrt((2hg)/(omega^(2)))~=20 cm.`