Analyzing the Motion of a Jumping Spring: Tips and Techniques

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In summary, the conversation discusses the motion of a spring when pushed against a wall or the ground. It is mentioned that the spring not only moves as a traveling object, but also contracts and expands. The speaker asks where they can find motion analysis on this topic and is directed to threads on the subject. They are advised to use conservation of energy in their analysis. The conversation then shifts to discussing the speed of the spring and when it will lose contact with the wall. The speaker expresses difficulty in solving the problem and is advised to start with simpler problems. They mention a video where a spring loses contact with a wall before reaching its normal length and ask for a solution using the equation mdu/dt=kx.
  • #1
luckis11
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A spring is pushed against the ground (or against the wall) and then left free to move. So, besides its motion as a traveling object, it also contracts and expands. Where can I find this motion analysis?
 
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  • #2
hi luckis11! :smile:
luckis11 said:
… Where can I find this motion analysis?

you'll find plenty of threads on it here …

use conservation of energy (kinetic energy + spring potential energy) :wink:​
 
  • #3
What do you mean "on it"? I'll find there the case of my question?
 
  • #4
luckis11 said:
What do you mean "on it"? I'll find there the case of my question?

similar cases :wink:

try conservation of energy … what do you get? :smile:​
 
  • #5
At what expansion will it lose contact with the wall?
 
  • #6
show us how far you've got, and where you're stuck, and then we'll know how to help! :smile:
 
  • #7
I am stuck here:

The speed of the spring (the spring=? I guess the centre of the mass of the spring) that will have just after losing contact with the wall is the last speed of the centre of the mass of the spring that it had just before it lost contact with the wall?
 
  • #8
This is actually a hard problem to solve, if you are talking about a "real" spring where the mass is distributed along the length of the spring.

I suggest you start with two easier problems:

1. A massless spring, with one end pushed against the ground and a mass on the other end.

2. The same as #1, but with equal masses on each end of the spring.
 
  • #9
You tell me to change my problem? That's a joke I guess.

Ι BET that it's not hard, but impossible.
 
  • #10
luckis11 said:
You tell me to change my problem? That's a joke I guess.
No, it's not a joke. If you don't know how to solve those easier problems, you don't know enough to understand a solution of your problem even if we told you how to do it.

If you can solve the second one correctly for a vertical spring and including gravity, then you will know WHY your problem is hard.

Ι BET that it's not hard, but impossible.
No, it's not impossible.
 
  • #11
A hand pushes it against the wall and then let it free. No mass attached.

At what expansion will it lose contact with the wall?
 
  • #12
luckis11 said:
A hand pushes it against the wall and then let it free. No mass attached.

At what expansion will it lose contact with the wall?

If the spring is horizontal it will lose contact with the wall when it is at its unstretched length, at time L/c, where L is the length and c is the axial wave velocity in the spring. [itex]c^2 = E/\rho[/itex] where [itex]E[/itex] is the elastic modulus and [itex]\rho[/itex] is the density (assuming the spring is "smeared out" into a uniform material)

If it is vertical the situation is more complicated because of the weight of the spring. When it leaves the ground it will not be uniformly stretched. If it is only compressed a small amount it may not leave the ground at all.

You haven't given us any clues about how much math and physics you know already, so I'm not going to try to explain why that is the answer.
 
  • #13
Where's that solution?

Should't there be a solution with the usual mdu/dt=kx? I think it should.
 
  • #14

FAQ: Analyzing the Motion of a Jumping Spring: Tips and Techniques

What is the purpose of analyzing the motion of a jumping spring?

The purpose of analyzing the motion of a jumping spring is to better understand the physics behind the movement of a spring when it is compressed and released. This can help in designing and improving spring-based devices such as shock absorbers, pogo sticks, and trampolines.

What are some tips for accurately measuring the motion of a jumping spring?

One tip is to use a high-speed camera to capture the movement of the spring. This will allow for more precise measurements and analysis of the motion. Another tip is to use a ruler or measuring tape to mark the starting and ending positions of the spring's movement. Additionally, it is important to make sure the surface the spring is jumping on is flat and even to avoid any interference with the motion.

How do you calculate the spring constant from the motion of a jumping spring?

The spring constant, also known as the stiffness coefficient, can be calculated by dividing the force applied to the spring by the displacement of the spring from its equilibrium position. This can be measured by using a force sensor and a displacement sensor. The formula for spring constant is k = F/x, where k is the spring constant, F is the force applied, and x is the displacement.

What factors can affect the motion of a jumping spring?

Some factors that can affect the motion of a jumping spring include the mass of the spring, the force applied to the spring, the surface the spring is jumping on, and the initial compression of the spring. Other factors such as air resistance and temperature can also have an impact on the spring's motion.

How can analyzing the motion of a jumping spring be useful in real-world applications?

Understanding the motion of a jumping spring can be useful in various real-world applications such as designing sports equipment, improving suspension systems in vehicles, and developing medical devices. It can also help in studying and predicting the behavior of other systems that exhibit similar motion, such as pendulums and oscillating objects.

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