A force of 60 N stretches a spring 30cm. A mass of 4 kg is hung from the spring and comes to rest.

It is then pulled down 10cm and released.

a) What is the period of this motion?

b) What are the magnitude and direction of the acceleration of the body when it is 5 cm above the equilibrium position, moving forward?

c)What is the tension in the spring when the body is 5cm above the equilibrium position?

d)What is the shortest time required to go from the equilibrium position to the point 5cm above?

e)If a small object were placed on the oscillating body would it remain in contact with the body or leave it?

f)If a small object were placed on the oscillating body and its amplitude doubled, where would the object and body begin to separate?

Explanation and work would help as well, Thanks!

Well.... I don't know what class this is for so lets start off with hooks law so we can find the spring constant.

k = F/x = (60N)/(0.3m) = :

Given:
F = ma
-kx = m(d^2x/dt^2)

0 = mx'' + cx' + kx
0 = x'' + (k/m)x
0 = x'' + (200 N/m)/(4 kg)x
0 = x'' + 50/s^2 x

Now just solve the equation and get:

x(t) = c1 cos(5 sqrt(2) t) + c2 sin(5 sqrt(2) t)

Impose x(0) = 0.07 m
x(0) = c1 + 0 = 0.07
c1 = 0.07

Impose v(0) = 0
v(0) = c2 (5 sqrt 2) = 0
c2 = 0;

So our spring equation looks like:

x(t) = 0.07cos(5 sqrt(2) t)

We will just approximate for the sake of simplicity from here on out:
x(t) = 0.07 cos(7.07 t) = 0
cos(7.07 t) = 0
7.07 t = 0 + 2 pi n, where n is an integer
t = 0.889 n

So that is our period is 0.889 s with our frequency 1.1254 Hz.

To figure out or magnitude and direction of the acceleration we need a couple derivatives.

x''(t) = -35. Cos(7.07 t]

The magnitude is 35, clearly.

So we need to know a time when it is at 5 cm.

set x(t) = 0.05m, solve for t.
I get t = 0.212034s.

x''(0.212034) = -2.5 m/s/s. And it is moving towards equilibrium.