# Hookes law question.

## Homework Statement

The scale of a spring balance that reads from 0 to 20.7 kg is 14.9 cm long. A package suspended from the balance is found to oscillate vertically with a frequency of 1.51 Hz. (a) What is the spring constant? (b) How much does the package weigh?

F=-kx

## The Attempt at a Solution

I know the answer, I'm not sure why it is done that way. You set kx=mg and solve for k, where x is the length of the spring. Why is that the x value? I thought the spring force only equaled gravity at the equilibrium point? Certainly the equilibrium point will change from the original length of the spring when you hang a mass from it.

Also, why is F=-kx some places, and others you just use F=kx, such as this problem?

gneill
Mentor

## Homework Statement

The scale of a spring balance that reads from 0 to 20.7 kg is 14.9 cm long. A package suspended from the balance is found to oscillate vertically with a frequency of 1.51 Hz. (a) What is the spring constant? (b) How much does the package weigh?

F=-kx

## The Attempt at a Solution

I know the answer, I'm not sure why it is done that way. You set kx=mg and solve for k, where x is the length of the spring.

What value for x are you going to plug in?

Also, why is F=-kx some places, and others you just use F=kx, such as this problem?

It depends upon the user's choice of coordinate system.

What value for x are you going to plug in?

It depends upon the user's choice of coordinate system.

I don't know what value of x I would use, but I'm still unsure why they give you that x value rather than the equilibrium x value. If the original spring with no mass on it hangs down a certain length, then how can that length be used to compute when the spring force equals gravity? That would mean the spring would not extend down.

gneill
Mentor
I don't see where they gave any x-value in the problem statement. And usually it's \Delta x that's involved (the displacement from the equilibrium position).

I don't know what value of x I would use, but I'm still unsure why they give you that x value rather than the equilibrium x value. If the original spring with no mass on it hangs down a certain length, then how can that length be used to compute when the spring force equals gravity? That would mean the spring would not extend down.

I think you were confused by the misleading infomration about the spring/balance being 149cm long. The solution to the problem only requires the mass and the frequency. No "x" involved.

I don't see where they gave any x-value in the problem statement. And usually it's \Delta x that's involved (the displacement from the equilibrium position).

Well in the solution manual they just use the original value of the spring, 14.9cm, as the x in kx=mg.

gneill
Mentor
Well in the solution manual they just use the original value of the spring, 14.9cm, as the x in kx=mg.

But the problem statement says that's the length of the scale, not the length of the spring. Also, the unloaded length of the spring wouldn't help; the unloaded length could be any length and it wouldn't tell you anything about its spring constant.

If they gave you the displacement of the center about which the oscillation occurs, that would be different.

I suspect that the question statement is incomplete or not properly posed.