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

• Danya314
In summary, the conversation is discussing how 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 problem involves using three types of energy and the given equation is v=A(k/m)^.5. The conversation also covers the initial and final positions, as well as the value of x at the equilibrium position. Ultimately, it is determined that x cannot be equal to zero at the equilibrium position and the formula for gravitational potential energy can be chosen as mgy with y = 0 at the equilibrium position.
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?

Danya314 said:

## 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?

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

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

Danya314
Would gravitational potential energy still be zero if I define xe to be 0?

Danya314 said:
Would gravitational potential energy still be zero if I define xe to be 0?

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

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 xe = 0. You should be able to determine xe 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 and AlephNumbers
I have it now. Thank you very much.

## 1. How do you calculate the velocity of a mass on a spring?

The equation for velocity of a mass on a spring is v = Aω cos(ωt + φ), where v is velocity, A is the amplitude of the oscillation, ω is the angular frequency, t is time, and φ is the phase constant.

## 2. What is the significance of the amplitude in the equation for velocity?

The amplitude, represented by the variable A, is the maximum displacement of the mass from its equilibrium position. It determines the maximum velocity that the mass will reach during its oscillation.

## 3. How does the angular frequency affect the velocity of a mass on a spring?

The angular frequency, represented by the variable ω, determines the speed at which the mass oscillates. A higher angular frequency will result in a higher velocity and a shorter period of oscillation.

## 4. Can the equation for velocity of a mass on a spring be used for any type of spring?

Yes, the equation v = Aω cos(ωt + φ) can be used for any type of spring, as long as the spring obeys Hooke's Law (F = -kx) and the mass is assumed to be negligible compared to the spring's mass.

## 5. How can the phase constant be determined in the equation for velocity of a mass on a spring?

The phase constant, represented by the variable φ, can be determined by using the initial conditions of the system. This can include the initial position and velocity of the mass, as well as the displacement and velocity at a certain point in time.

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