Mass-spring system in SHM concept question

[freshcoast]

Homework Statement

Homework Equations

The Attempt at a Solution

Well since spring is in SHM, only conservative forces are at play here. So using conservation of energy, the kinetic energy would be the change in potential energy. Which I have set up as

KE = 1/2k(A^2)² - 1/2k(A/4)²

after some algebra, combining like terms with common denominators I am left with,

KE = (15/32)kA²

now when I take the fraction of the two, PE/KE of

(15/32)kA² / 1/2mv²

I get (15/16)kA²/mvf²

So I'm thinking my answer would be 15/16

I don't know whether to just ignore the variables since there was no value in them or I find a way to relate them since I was given a frequency?

 

[collinsmark]

Homework Statement

Homework Equations

The Attempt at a Solution

Well since spring is in SHM, only conservative forces are at play here. So using conservation of energy, the kinetic energy would be the change in potential energy. Which I have set up as

KE = 1/2k(A^2)² - 1/2k(A/4)²

after some algebra, combining like terms with common denominators I am left with,

KE = (15/32)kA²

So far so good.

now when I take the fraction of the two, PE/KE of

(15/32)kA² / 1/2mv²

I get (15/16)kA²/mvf²

Sorry but I'm not following you there.

What's the total energy of the system? Once you find that you can find the ratio of kinetic energy over total energy.

So I'm thinking my answer would be 15/16

It might be 15/16. But if so, you haven't quite shown why yet.

I don't know whether to just ignore the variables since there was no value in them or I find a way to relate them since I was given a frequency?

Hint: your next step is to find the total energy of the system.

[Edit: Another hint: the total energy of the system is conserved, and doesn't change with time. ]

 

[HallsofIvy]

The reason this is a "conceptual" problem, as opposed to a "calculation" problem is that you can use the fact that the potential energy of a spring is proportional to the extension or compression of the spring to quickly find the ratio. When the mass is at distance A from the equilibrium all of the energy is potential, when it is distance 0 from equilibrium, none of it is. when it is at distance A/4 from equilbrium (and so 3A/4 from greatest extension) what fraction of the energy is potential energy? What fraction is kinetic?

 

[freshcoast]

Oh I think I've figured it out. The total energy of the system can either be when x is at it's max "A", or when the velocity is at it's max = v_max. So I need an equation for the kinetic energy, which I set up as

1/2mvf^2 = 1/2kA^2, which gives me the Vfinal I can use to substitute into the kinetic energy equation above which allows me to take the ratio, and everything being canceled out I am just left with 15/16

 

[collinsmark]

[...] the potential energy of a spring is proportional to the extension or compression of the spring [...]

I think you mean the potential energy is proportional to the square of the extension or compression.

Oh I think I've figured it out. The total energy of the system can either be when x is at it's max "A", or when the velocity is at it's max = v_max. So I need an equation for the kinetic energy, which I set up as

1/2mvf^2 = 1/2kA^2, which gives me the Vfinal I can use to substitute into the kinetic energy equation above which allows me to take the ratio, and everything being canceled out I am just left with 15/16

I am not familiar with whatever you are doing with the "1/2mvf^2." Maybe you mean [itex] \frac{1}{2}m(v_{max})^2?[/itex].

But yes, the potential energy of the system is [itex] \frac{1}{2}kA^2 [/itex] when the extension is equal to A. You can use that (and a hints from HallsofIvy and myself [my previous post]) to determine the total energy of the system.

In other words, you could solve for [itex] v_{max} [/itex] and substitute things around. But it's not necessary to even bring [itex] v_{max} [/itex] into the problem. If you know what the potential energy is at maximum extension, then you also know the total system energy.