Spring oscillation equilibrium point

In summary, a 1.70 kg tray attached to a vertical spring with a force constant of 180N/m has a 295 g metal ball inside. The tray is pushed down 16.5 cm and released from rest. The height above point A when the ball leaves the tray is 27.4 cm. To find the time elapsed between releasing the system and the ball leaving the tray, the normal force exerted on the ball by the tray should be considered. Acceleration as a function of time is also necessary.
  • #1
Julie323
24
0

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!
 
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  • #2
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.
 
  • #3
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?
 
  • #4
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?
 
  • #5
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!
 
  • #6
Are you sure this is introductory physics questions? I have seen this in university freshman physics, hahaha
 
  • #7
Freshman physics is introductory physics. :wink:

Hint: Think about the normal force on the ball exerted by the tray.
 
  • #8
Oh, that would save all the graphing. But you still need acceleration as a function of time.
 

1. What is the spring oscillation equilibrium point?

The spring oscillation equilibrium point is the position at which the spring is at rest, with the force of gravity balanced by the restoring force of the spring. This is also known as the mean position or the equilibrium position.

2. How is the equilibrium point determined for a spring?

The equilibrium point for a spring can be determined by finding the point at which the net force on the spring is zero. This can be calculated by setting the equations for the spring force and the gravitational force equal to each other and solving for the position.

3. What factors affect the equilibrium point of a spring oscillation?

The equilibrium point of a spring oscillation can be affected by several factors, including the stiffness of the spring, the mass of the object attached to the spring, and the force of gravity.

4. How does the equilibrium point change with different spring constants?

The equilibrium point of a spring oscillation will change with different spring constants. A stiffer spring will have a higher equilibrium point, while a more flexible spring will have a lower equilibrium point.

5. Can the equilibrium point of a spring oscillation be changed?

Yes, the equilibrium point of a spring oscillation can be changed by altering the factors that affect it, such as the spring constant or the mass attached to the spring. It can also be changed by applying an external force to the spring.

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