Spring oscillation equilibrium point

[Julie323]

Homework Statement

A 1.70 kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 180N/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 16.5 cm below its equilibrium point (call this point A) and released from rest.

How high above point will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.)

Homework Equations

F=-kx

The Attempt at a Solution

I was initially thinking that the ball would leave the tray when the spring was at its uncompressed length. I solved for how far the spring was compressed at its equillibrium point. I used F=kx. (1.7+.295)(9.8)=(180)(x) This gave me a value of x of .109 m. This was wrong, any help would be greatly appreciated!

 

[Delphi51]

Wow, this is the most interesting problem I've seen for a while!
I can't quite put my finger on the condition that causes the ball to float but I suspect it has something to do with acceleration. Recommend you find the frequency of oscillation, then write the formula for the stretch as a function of time, differentiate it to get the velocity and acceleration formulas. The formulas will fail when the ball floats (mass changes) but they should be good right up to the time we are interested in! It will be interesting to graph x, v and a for a period.

 

[Julie323]

okay, well I found the angular frequency to be 9.5 rad/s using sqrt{k/m}. I then found equations: x=16.5cos(9.5t), v=-9.5*16.5sin(9.5t), and a=-90.25*16.5cos(9.5t). Where do I go from here?

 

[Julie323]

I just figured part A out. I thought the problem was saying A was the equillibrium point, but it is actually the point when it is pushed down 16.5 cm, so the answer was 10.9+16.5 or 27.4 cm. Now part B is giving me trouble:

How much time elapses between releasing the system at point and the ball leaving the tray?

I tried putting 27.4 in for x and solving for t in the equation x=16.5cos(9.5t) but either the equation is wrong or something else is messed up because that give cosine a value of greater than 1. Any ideas?

 

[Delphi51]

What part A? I see only the question about the point where the ball leaves the tray. I take the 16.5 cm to be the initial value of stretch. Xo = .165 m. So x = .165 cos(ωt). Agree ω = 9.5.
I still think you need to sketch some graphs of x, v and a to spot the condition where the ball departs from the pan!

 

[yinx]

Are you sure this is introductory physics questions? I have seen this in university freshman physics, hahaha

 

[vela]

Freshman physics is introductory physics.

Hint: Think about the normal force on the ball exerted by the tray.

 

[Delphi51]

Oh, that would save all the graphing. But you still need acceleration as a function of time.