How Do You Solve an Oscillation Problem Involving a Vertical Spring and Block?

[kittykatxox]

a 850 g block is attached to a vertical spring whose stiffness constant is 9N/mThe block is released at the position where the spring is unextended. a) What is the maximum extension of the spring? b) How long does it take the block to reach the lowest point?

so I found w by using k=mw^2
9=0.85w^2
w=3.26 rad/s
next I found the max extension by using F=-(mw^2)x
(0.85*9.8)= -(0.85*(3.26)^2)x
x=0.93
I'm not sure what to do next, any help is appreciated

Thanks

 

[collinsmark]

so I found w by using k=mw^2
9=0.85w^2
w=3.26 rad/s

Your ω looks pretty good. (Except for a minor rounding error. You might wish to check your significant figures.)

next I found the max extension by using F=-(mw^2)x
(0.85*9.8)= -(0.85*(3.26)^2)x
x=0.93

Sorry, but F = -kx, alone, isn't going to work for this problem all by itself. (Although you still might find it useful, so keep it in your back pocket.)

F = -kx would work if you were trying to find the distance between the old equilibrium position, before the mass was attached, to the new equilibrium position after the mass is attached.

But in this problem, the mass doesn't simply fall to the new equilibrium position and stay there. Instead it falls past the new equilibrium position and keeps going. Eventually, its velocity drops to zero for an instant, but it doesn't stop there! The spring then pulls the mass back up, again past the new equilibrium position, and eventually the mass stops back up at the top where it was initially released. The whole process then repeats. The block has gone into simple harmonic motion.

There are two ways to determine the distance from the highest position to the lowest position the mass reaches.
a. The first way is this: you could describe the geometry of this motion, and then deduce the peak-to-peak difference from that.
b. The second way is to use conservation of energy. Consider the two places in the oscillation where the velocity is zero (once at the top and once at the bottom). At both of these places, the kinetic energy of the mass is zero. Use conservation of energy, considering the gravitational potential energy, and the potential energy stored in the spring.

I'm not sure what to do next, any help is appreciated
Thanks

Again, you'll have to redo your calculation of x, as I've discussed above. But you do have a good value for ω already! (Except for the minor rounding error.)

If you know ω, can you calculate the period, T of oscillation? If the mass starts at the top, what fraction of T does it take to get to the bottom?