Calculating Effective Spring Constant for Platform Supported on 4 Springs

In summary: In other words, the kinetic energy of the clay and platform system at the maximum amplitude of the oscillation should be equal to the kinetic energy of the clay alone just before collision.In summary, the platform of mass 0.8 kg is supported on four springs (only two springs are shown in the picture, but there are four altogether). A chunk of modeling clay of mass 0.6 kg is held above the table and dropped so that it hits the table with a speed of v = 0.9 m/s. The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 6 cm below its original position.
  • #1
huskydc
78
0
A platform of mass 0.8 kg is supported on four springs (only two springs are shown in the picture, but there are four altogether). A chunk of modeling clay of mass 0.6 kg is held above the table and dropped so that it hits the table with a speed of v = 0.9 m/s.

The clay sticks to the table so that the the table and clay oscillate up and down together. Finally, the table comes to rest 6 cm below its original position.

-->What is the effective spring constant of all four springs taken together?

i did the following: F = kx

so F/x = k

i'm thinking the only force contributing to this is dropping (force) of the modeling clay on the platform. or does the weight (gravity) of the platform contribute to any of this?
 
Physics news on Phys.org
  • #2
Presumably, the mass of the platform is considered in the initial position of the springs.

The additional mass of the clay displaces the springs further.

At rest, this becomes a static problem.

If one is considering the oscillation, then one would have to consider the combined mass.
 
  • #3
If the table has come to rest, then the velocity with which the clay struck the table and any resulting oscillations are irrlevant.
What you did is pefectly correct. Also, the weight of the platform doesn't get involved since we are told it stopped (6 cm) below its original position - so it's only the 6 cm compression that has to be considered and all of that is provided by the (static force) of the weight of the clay on the platform/spring combination.
 
  • #4
The only thing that is relevant here is that you are told the mass of the clay and that the table comes to rest 6 cm. below its original height. In other words, compressing the spring 6 cm= 0.06 m will support a force of 0.8g = 0.8(9.81) Newtons.

The fact that clay was originally dropped from a height affects the oscillation but not the final equilibrium position. The mass (and weight) of the table itself determines the original height of the table. It is the added weight of the clay that causes the 6 cm compression.
 
  • #5
ok, the figured out the k = 98.1 N/m,

next the problem asks:

With what amplitude does the platform oscillate immediately after the clay hits the platform?

i first used the momentum conservation, and found the velocity of the clay+platform system is equal to .36

since total E = .5kA^2

i'm not sure here, but i set initial E = final E like the following:

initial KE + initial PE = final KE + final PE
.5mv^2 + 0 = .5(m of clay + M of platform).36^2 + .5kx^2

then the final KE equal to .5kA^2, it got me the incorrect answer, but is there anything wrong with this though?
 
  • #6
At maximum compression (the amplitude of the oscillation), the speed of the clay and table is 0, not ".36" (that's the speed of clay and table immediately after impact which is not relevant).
 
  • #7
well, the book has pretty much the same example, it says to find the final speed of the clay and platform system, find kinetic energy of the system after collision then set this KE to .5kA^2, tried that example myself, it didnt work out ...
 
  • #8
But you didn't do that, did you? You had ".5mv^2 + 0 = .5(m of clay + M of platform).36^2 + .5kx^2" That ".4kx^2" is the ".5k A^2" but you haven't set that equal to the kinetic energy after collision. The kinetic energy of the clay, platform combination immediately after collision should be exactly the same as the kinetic energy of the clay alone just before collision- that's what I was using.
 

Related to Calculating Effective Spring Constant for Platform Supported on 4 Springs

1. How do I calculate the effective spring constant for a platform supported on 4 springs?

To calculate the effective spring constant for a platform supported on 4 springs, you will need to know the individual spring constants of each spring as well as the arrangement of the springs (parallel or series). The formula for calculating the effective spring constant is 1/keff = 1/k1 + 1/k2 + 1/k3 + 1/k4, where keff is the effective spring constant and k1, k2, k3, k4 are the individual spring constants.

2. What is the significance of calculating the effective spring constant for a platform supported on 4 springs?

Calculating the effective spring constant allows you to determine the overall stiffness of the platform, which is important for understanding how it will respond to applied forces. This information can be useful in designing and optimizing the platform for its intended use.

3. Can the effective spring constant be negative?

No, the effective spring constant cannot be negative. This is because springs always exert a restoring force in the opposite direction of its displacement, which results in a positive spring constant value.

4. How does the arrangement of the springs affect the effective spring constant?

The arrangement of the springs (parallel or series) affects the effective spring constant by changing the overall stiffness of the system. In parallel arrangement, the effective spring constant will be larger than any individual spring constant, while in series arrangement, the effective spring constant will be smaller than any individual spring constant.

5. Can the effective spring constant change over time?

In most cases, the effective spring constant will remain constant over time. However, if the individual springs experience wear and tear or changes in temperature, the effective spring constant may change slightly. It is important to regularly check and maintain the individual springs to ensure the accuracy of the calculated effective spring constant.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
908
  • Introductory Physics Homework Help
Replies
17
Views
398
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
410
  • Introductory Physics Homework Help
Replies
3
Views
894
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
6K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Replies
2
Views
5K
Back
Top