Why Does the Mass on a Spring Homework Yield Contradicting Equations?

In summary, the conversation discusses the placement and release of a mass on a vertically placed spring, and the resulting equations for force and energy. It also addresses a question about firing a mass horizontally at a wall-mounted spring and the resulting maximum displacement. The experts provide explanations for these scenarios and confirm the correct equations and understanding.
  • #1
rammer
23
0
Suppose we have a spring placed vertically on a table. We put a body with mass "m" on it. The spring compresses by "x".
Then sum of forces is zero, therefore:

kx=mg, so k=mg/x

But if we look on it as the energy problem, then: we hold the body on the string, then we release it, so loss of gravitational potential energy is "mgx" and gain of spring energy is 0.5*k*x^2.
From that:

k=2mg/x, which contradicts previous equation.

Where I did a mistake?
 
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  • #2
rammer said:
Suppose we have a spring placed vertically on a table. We put a body with mass "m" on it. The spring compresses by "x".
Then sum of forces is zero, therefore:

kx=mg, so k=mg/x
Here you have gently lowered the mass onto the spring. Once placed in its new equilibrium position (where the spring is compressed by an amount x = mg/k), it just sits there. Note that your hand does work (negative work) on the mass as you lower it, reducing the total energy of the system.
But if we look on it as the energy problem, then: we hold the body on the string, then we release it, so loss of gravitational potential energy is "mgx" and gain of spring energy is 0.5*k*x^2.
From that:

k=2mg/x, which contradicts previous equation.

Where I did a mistake?
Here you just release the mass. So all the initial gravitational PE remains to be converted to spring PE and kinetic energy, with none being removed by your hand. When it gets to the equilibrium position (at x = mg/k) it still has kinetic energy left, so it keeps going. It reaches the point x2 = 2mg/k, where the KE is momentarily zero, then comes back up. It will continue to oscillate between the two extreme positions.
 
  • #3
This is a well known question dealing with springs !
When you lower a weight onto a spring the force exerted on the spring increases from 0 up to the maximum (the weight of the object).
Therefore the work done compressing the spring = AVERAGE force x distance
= 0.5F x distance.
During the placing of a weight on a spring your hand is part of the process and ensures that the force on the spring increases from 0 to the maximum.
On the other hand (sorry !) if you release (drop) the weight then the force on the spring is constant (mg) for the compression and the extra (0.5mgh) energy appears as KE... the mass on the spring will bounce up and down

Doc Al has said the same !
 
Last edited:
  • #4
Thanks for both explanations :)
 
  • #5
I'd also like to thank Doc Al and technician for the answers. I missed this thread last month when I asked nearly the same question:
https://www.physicsforums.com/showthread.php?t=576713

Incidentally, I received a not-as-helpful answer from the askthephysicist.com site when I inquired there:

Your question is meaningless, the old "comparing apples and oranges" thing.
The x for the carefully placed mass is the equilibrium position, the x for the
dropped mass is not.

I think I get that now, thanks to the extra explanation given in this thread. If I might add a question:

I understand that when the mass is placed on the spring, displacement x = mg/k. When the mass is dropped on the spring, (non-equilibrium) displacement x = 2mg/k.

Per my question in the other thread, if we instead fired the mass horizontally at a wall-mounted spring… we can no longer talk of g, so have to look at the projectile's KE in terms of its mass and velocity. Doing so, we get: x = √(m/k) * v (momentarily, anyway, before the spring completely pushes the mass away again)

Is that correct? The confirmation would be greatly appreciated. (The difference between x proportionate to m/k when the mass is placed or dropped onto the spring, vs proportionate to √(m/k) when it's fired at the side-mounted spring, is what had me confused… but I believe I now see the misunderstanding that led to my confusion.)
 
  • #6
superalias said:
Per my question in the other thread, if we instead fired the mass horizontally at a wall-mounted spring… we can no longer talk of g, so have to look at the projectile's KE in terms of its mass and velocity. Doing so, we get: x = √(m/k) * v (momentarily, anyway, before the spring completely pushes the mass away again)
Yes, that's correct. (Assuming no energy is lost in the collision.) That expression gives the maximum displacement from the initial position.
 
  • #7
Thanks for the confirmation!

Of course, an answer here just leads to more questions there… but that's for other threads down the road. Thanks again!
 

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

The equation for the period of a mass on a spring is T = 2π√(m/k), where T is the period (in seconds), m is the mass (in kilograms), and k is the spring constant (in N/m).

2. How does the mass affect the period of a spring?

The period of a spring is directly proportional to the square root of the mass. This means that as the mass increases, the period also increases. This relationship can be seen in the equation T = 2π√(m/k), where a larger mass will result in a longer period.

3. What is the relationship between the spring constant and the period of a mass on a spring?

The spring constant and the period of a mass on a spring have an inverse relationship. This means that as the spring constant increases, the period decreases. This can be seen in the equation T = 2π√(m/k), where a larger spring constant will result in a shorter period.

4. How does the amplitude affect the period of a mass on a spring?

The amplitude does not affect the period of a mass on a spring. The period is only dependent on the mass, spring constant, and gravitational acceleration. The amplitude only affects the maximum displacement of the mass from its equilibrium position.

5. How does the initial displacement of the mass affect the period of a spring?

The initial displacement of the mass does not affect the period of a spring. The period is only dependent on the mass, spring constant, and gravitational acceleration. The initial displacement only determines the starting position of the mass on the spring, but does not affect the overall period of the motion.

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