How to find the equation for velocity of a mass on a spring?

[Danya314]

Homework Statement

The problem is to use conservation of energy to determine the equation for the velocity at the equilibrium position of an oscillating mass on a vertical spring. The question says to use three types of energy. The question gives the equation, the problem is to solve for it. The equation is v=A(k/m)^.5

Homework Equations

Uelastic+Ugravitational+Ukinetic=Uelastic+Ugravitational+Ukinetic

The Attempt at a Solution

The initial position I chose was at the bottom of the mass's motion and the final at equilibrium, so inital kinetic and final elastic and gravitational are zero.

mgh+.5kx^2=.5mv^2
since h and x are the amplitude,
mgA+.5kA^2=.5v^2
simplifying gives,
2gA+(k/m)A^2=v^2
How do I get rid of the gravity term?

 

[TSny]

The Attempt at a Solution

The initial position I chose was at the bottom of the mass's motion and the final at equilibrium, so inital kinetic and final elastic and gravitational are zero.

mgh+.5kx^2=.5mv^2
since h and x are the amplitude,
mgA+.5kA^2=.5v^2

If you are taking gravitational PE to be zero at the equilibrium position, then what should be the sign of the gravitational PE at the bottom of the motion?

x represents the amount that the spring is stretched from the unstretched length. If the mass hangs at rest at the equilibrium position, then the spring is stretched. So, x is not equal to zero at the equilibrium position. Also, x is not equal to A at the bottom of the motion.

 

[Danya314]

Thanks for the help. But why wouldn't x be amplitude at the bottom?

 

[TSny]

The spring already has some stretch at the equilibrium position. Let this stretch be x_e.

Then, at the bottom the spring will be stretched by x = x_e + A.

 

[Danya314]

Would gravitational potential energy still be zero if I define xe to be 0?

 

[AlephNumbers]

Would gravitational potential energy still be zero if I define xe to be 0?

That depends. Are you redefining the axis so that X_e is the new zero?

 

[TSny]

The formula PE = 1/2 k x² for the spring assumes that x is the amount of stretch of the spring. Since the spring is stretched at the equilibrium position, you cannot let x_e = 0. You should be able to determine x_e in terms of m, g, and k. (Think about the mass hanging at rest at the equilibrium position.)

For the gravitational PE you can use PE = mgy and you can choose y = 0 anywhere you want. So, you can choose y = 0 at the equilibrium position if you want. Then PE for gravity = 0 at the equilibrium position.

 

[Danya314]

I have it now. Thank you very much.