Spring and gravitational energies (Simple harmonic motion)

[ichivictus]

Problem:

Show that the combined spring energy and gravitational energy for a mass m hanging from a light spring of force constant k can be expressed as 1/2 ky², where y is the distance above or below the equilibrium position.

Figure shows a block connected to spring, where equilibrium is, and where the spring's unstretched position is. The distance from unstretched position to equilibrium is defined as h. From equilibirum to mass is y.


The answer starts off as

U = 1/2 k(h+y)² - mg((h/2) +y)

From there it says mg=kh and substitutes for mg then solves.

I'm confused how U equals all that. I get how potential energy of a spring is 1/2 kx^2 for a spring or mgh in other cases, but this is quite confusing.

Then where did the 1/2 from h/2 come from?

 

[paisiello2]

The answer I can think of is that all energy is relative. So if the datum is conveniently chosen so that some variables drop out then that is still valid.

As an alternate approach, why don't you assume a datum distance of "d" and assume the two terms can be set to ky^2. Then solve for d.

 

[Simon Bridge]

The potential energies are relative - but the position the values are relative to is provided as part of the problem setup.
(Well - sort of...)

The idea is that the equilibrium position is defined to be where the force of gravity is the same as the force of the spring ... since gravity is a constant, and the spring is extended by length h, this means: kh=mg

To understand the original equation, first sketch out what they are describing and then work out the energies, and add them up.

 

[paisiello2]

The potential energies are relative - but the position the values are relative to is provided as part of the problem setup.

Nonsense. The gravitational potential energy datum is not given at all in the set up.

 

[HalfLostAndSearching]

I would like to clarify and explain the concept of spring and gravitational energies in simple harmonic motion.

First, let's define some terms:

- Spring energy: This is the potential energy stored in a spring when it is stretched or compressed from its equilibrium position.
- Gravitational energy: This is the potential energy due to the position of an object in a gravitational field.

In the given problem, we have a mass m hanging from a light spring with force constant k. The equilibrium position is where the spring is neither stretched nor compressed, and the distance from this position to the mass is defined as y.

Now, let's analyze the equation provided:

U = 1/2 k(h+y)2 - mg((h/2) +y)

The first term, 1/2 k(h+y)2, represents the potential energy stored in the spring. This can be derived from the equation for spring energy, 1/2 kx^2, where x represents the displacement from the equilibrium position. In this case, the displacement is (h+y), as the spring is already stretched by a distance h and the mass adds an additional displacement of y.

The second term, - mg((h/2) +y), represents the gravitational potential energy of the mass. The negative sign is due to the fact that the gravitational force is acting in the opposite direction of the displacement. The term (h/2) represents the distance from the equilibrium position to the center of mass of the mass, which is h/2 below the equilibrium position. The additional y represents the distance from the center of mass to the actual position of the mass.

Now, the problem states that mg = kh, which means that the gravitational force is equal to the force exerted by the spring at the equilibrium position. This is because at equilibrium, the net force acting on the mass should be zero.

Substituting this value in the second term, we get:

U = 1/2 k(h+y)2 - kh((h/2) +y)

= 1/2 k(h+y)2 - 1/2 kh(h+y)

= 1/2 k(h+y)(h+y-h)

= 1/2 kh(h+y)

= 1/2 ky2

This is the expression we were asked to show, where y represents the distance above or below the equilibrium position.

To summarize, the combined spring and gravitational energies for