A2.00 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 190 n/m and a 295 g metal ball is in the tray. the spring is below the tray, so it can oscillate up-and-down. the tray is then pushed down 17.4 cm below its equilibrium point (call this point a) and released from rest.
(a) how high above point a will the tray be when the metal ball leaves the tray? (b) how much time elapses between releasing the system at point a and the ball leaving the tray? (c) how fast is the ball moving just as it leaves the tray?

a)

b)

c)

Explanation:

Given:

(a)

Now the ball will leave the tray at the highest oscillation position from the equilibrium where the spring force becomes equal to the combined weights.

mathematically:

where:

x = maximum position of ball and tray from the initial mean position.

∴Distance from point A which is fully compressed position:

(b)

Simple harmonic motion as a function of time is given as:

....................................(1)

where:

angular frequency

t = time

phase angle= 0 (initially at t=0 the spring is fully compressed)

y = displacement

Now from equation (1):

...........................(2)

we know

Now, putting reaspective values in eq. (2)

(c)

We have the equation for total mechanical energy of SHM as:

it has to be b, about 97 percent.

i think the answer is a, hope this : )