plsss help meee ..?

Question

question 1 :
- A 29.0 kg block at rest on a horizontal frictionless air track is connected to the wall via a spring. The equilibrium position of the mass is defined to be at x=0. Somebody pushes the mass to the position x= 0.350 m, then lets go. The mass undergoes simple harmonic motion with a period of 4.50 s. What is the position of the mass 3.600 s after the mass is released?

- Consider the same mass and spring discussed in the previous problem. What is the magnitude of the maximum acceleration the mass undergoes during its motion?
the figure : http://www3.0zz0.com/2006/12/22/10/61592534.jpg
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question 2 :
A 8.40 kg mass suspended from a spring with spring constant, k = 800.0 N/m, extends it to a total length of 0.270 m. Find the total length of the spring when a 13.40 kg mass is suspended from it.

I begin to solve it's problem but i don't know how to find the total length of the spring !!
F=mg=13.40*9.81=131.454 N
x=F/k=131.454/800=0.1643m

Answer 1

1. Since the track is frictionless (I assume it's free of air drag too) it will continue to oscillate with an amplitude of 0.35 m, both plus and minus. Simple harmonic motion says you can describe its position with amplitude and a sine function:
x = 0.35 * sin(90+[360*t/4.5]) m
Notice that if t = 0, then x = 0.35 * sin(90) m = 0.35 m,
if t = 4.5/4, then x = 0.35 * sin(90+90) m = 0 m
if t = 4.5/2, then x = 0.35 * sin(90+180) m = -0.35 m
I hope you agree that that's what it should be at those times. So it's working. Plug in t = 3.6 s

part a. If you know calculus, take the 2nd derivative.

Another way: max acceleration will happen when the spring's force is max. And that's at max deflection from the rest position: x = + or - 0.35 m. The force and therefore the acceleration is 0 when x=0. And the time to go between these positions is 4.50/4 s. The formula
x = (1/2)*a*t^2
gives you the average acceleration over time t. The force decreases linearly from max to zero as it moves from x = +0.35 to x = 0. So the acceleration at max is twice the average.

2. The work you did gives you the length of extension when you hang 13.40 kg on it. Call that result L2. You need to do the same calculation with the 8.40 kg mass. Call that result L1. The total extension (0.270 m is given) is it's normal, no load, length plus L1, how much the first mass stretched it. So when no load is on it, the length is 0.27 m - L1 m. Call that Ls.

Now, Ls + L2 should give you the result you need.

Answer 2

3

Answer 3

Do a little research under Hooke's Law. Seriously.

Answer 4

i'm going to have nightmares.