Gregory’s Classical Mechanics Mass on a spring

In summary, the mass on a spring is compressed by a/3 when it is lifted to a height 3a/2 above the floor.
  • #1
blalien
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0
[SOLVED] Mass on a spring

Homework Statement


This problem is from Gregory’s Classical Mechanics

A light spring of natural length a is placed on a horizontal floor in the upright position. When a block of mass M is resting in equilibrium on top of the spring, the compression of the spring is a/15. The block is now lifted to a height 3a/2 above the floor and released from rest. Find the compression of the spring when the block first comes to rest.

Homework Equations


Mgh = the potential energy of the mass at a height h above the floor
1/2kx^2 = the potential energy of the spring, since we are assuming that the spring obeys Hooke's Law

The Attempt at a Solution



It doesn’t help that the book never defines “natural length” or “compression.” But anyway, this is my logic:

Assume that, on the horizontal floor, V = 0. So, you have two systems and two equations:

Mass starts at rest on spring -> Mass compresses spring by a/15
Mga = Mg(14a/15)+1/2k(a/15)^2

Mass starts at 3a/2 above floor -> Mass compresses spring by x
Mg(3a/2) = Mg(a-x)+1/2kx^2

Then you solve for k and x. Unfortunately, this yields the wrong answer, which is apparently x = a/3. I don’t want to be told exactly how to do the problem. But, could you please just point to the faulty part in my logic? I would really appreciate the help.

Thanks!
 
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  • #2
blalien said:
It doesn’t help that the book never defines “natural length” or “compression.”
Natural length = uncompressed/unstretched length; compression = distance the spring is compressed (away from its uncompressed position).

Mass starts at rest on spring -> Mass compresses spring by a/15
Mga = Mg(14a/15)+1/2k(a/15)^2
Use the given information to determine the spring constant.

Mass starts at 3a/2 above floor -> Mass compresses spring by x
Mg(3a/2) = Mg(a-x)+1/2kx^2
Measure spring energy from the unstretched position. When the spring is pulled up to 3a/2, what is its spring potential energy? (Not zero!)
 
  • #3
Oh, I see. So, the mass and the spring are lifted up to the height 3a/2. I was assuming that the mass alone was lifted up then dropped, but I guess your way makes more sense. Then the second equation is actually:

Mg(3a/2)+1/2k(a/2)^2 = Mg(a-x)+1/2kx^2

That still doesn't yield the right answer, though.

So, when there is no mass on the spring, the spring's equilibrium position is clearly at height h=a. Therefore, the spring energy is 1/2k(h-a)^2, where h is the height of the top of the spring.

When the mass is on the spring, does this remain true? Is the equilibrium position still at h=a, or is it at h=14a/15? Or do we have to redefine the spring energy of the system?
 
  • #4
blalien said:
Oh, I see. So, the mass and the spring are lifted up to the height 3a/2. I was assuming that the mass alone was lifted up then dropped, but I guess your way makes more sense.
Now that I reread the problem, I think your interpretation makes more sense!

First find the spring constant using the initial information as I pointed out earlier, then apply the second equation you had before:
blalien said:
Mass starts at 3a/2 above floor -> Mass compresses spring by x
Mg(3a/2) = Mg(a-x)+1/2kx^2

This should work. (Sorry about that!)
 
  • #5
Well yeah, therein lies the problem.

So we start with Mga = Mg(14a/15)+1/2k(a/15)^2. Solving for k gives k = 30Mg/a

Then we have Mg(3a/2) = Mg(a-x)+1/2kx^2
Mg(3a/2) = Mg(a-x)+1/2(30Mg/a)x^2

Then, solving for x gives x = a(sqrt(31)+1)/30. The solution given in the back of the book is, "Spring is compressed by a/3." Now, the book's solutions have been wrong before, but x = a/3 does sound more intuitive. So one of those equations is incorrect. The question is finding the error.
 
  • #6
blalien said:
Well yeah, therein lies the problem.

So we start with Mga = Mg(14a/15)+1/2k(a/15)^2. Solving for k gives k = 30Mg/a
To find the spring constant, use a force equation, not an energy one. You are told that a certain force (Mg) compresses the spring by a certain amount (a/15) . Use Hooke's law to find k.
 
  • #7
Ah, there we go. Thanks so much!
 

Related to Gregory’s Classical Mechanics Mass on a spring

1. What is Gregory's Classical Mechanics Mass on a spring?

Gregory's Classical Mechanics Mass on a spring is a simple physics problem that involves a mass attached to a spring that is oscillating up and down. It is used to teach fundamental principles of classical mechanics such as Hooke's law and simple harmonic motion.

2. What is Hooke's law?

Hooke's law states that the force exerted by a spring is directly proportional to the amount of stretch or compression of the spring. This means that the more the spring is stretched or compressed, the greater the force it exerts. In mathematical terms, this can be written as F = -kx, where F is the force, k is the spring constant, and x is the displacement from equilibrium.

3. How does the mass on a spring system exhibit simple harmonic motion?

A mass on a spring system exhibits simple harmonic motion because the force exerted by the spring is directly proportional to the displacement of the mass from its equilibrium position. This results in a repeating pattern of motion where the mass moves back and forth around the equilibrium position with a constant period and amplitude.

4. What is the equation for the period of a mass on a spring system?

The period of a mass on a spring system can be calculated using the equation T = 2π√(m/k), where T is the period, m is the mass of the object, and k is the spring constant. This equation shows that the period is independent of the amplitude of the motion, but dependent on the mass and spring constant.

5. How does the motion of a mass on a spring system change with changes in mass or spring constant?

If the mass or spring constant is changed in a mass on a spring system, it will affect the period and amplitude of the motion. Increasing the mass will result in a longer period and smaller amplitude, while increasing the spring constant will result in a shorter period and larger amplitude. However, the fundamental principles of simple harmonic motion and Hooke's law will still apply.

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