- #1

Asad Raza

- 82

- 3

Additiinally, How do we determine the rate of change of potential energy in such case?

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- Thread starter Asad Raza
- Start date

In summary, the rate of change of potential energy is always the same as the rate of change of kinetic energy in a mass spring system. Additionally, to determine the rate of change of potential energy, you need to differentiate the two rates of change and compare them.

- #1

Asad Raza

- 82

- 3

Additiinally, How do we determine the rate of change of potential energy in such case?

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- #2

ShayanJ

Gold Member

- 2,810

- 605

To determine the rate of change of potential energy:

Step1) Consider the most general trajectory for a harmonic oscillator:## x(t)=A\cos \omega t+B\sin\omega t ##

Step2) Put ##x(t)## in ## V(t)=\frac 1 2 kx(t)^2=\frac 1 2 m\omega^2 x(t)^2 ##.

Step3) Differentiate ## V(t) ## w.r.t. time.

Step4) Differentiate ## K(t) ## w.r.t. time.(you calculated it here!)

Step5) Compare the results from Step3 and Step4.

- #3

TJGilb

- 156

- 38

##\frac 1 2 m(A \omega sin(\omega t))^2+mg(h-Acos(\omega t))+\frac 1 2 k(Acos(\omega t))^2=constant##

Taking the rate of change with respect to time we get

##mA^2 \omega^3 sin(\omega t)cos(\omega t)+mgA \omega sin(\omega t)-kA^2 cos(\omega t)sin(\omega t)=0##

Putting the rate of change of the potential energies on one side and the kinetic energy on the other, we get

##mA^2 \omega^3 sin(\omega t)cos(\omega t)=kA^2 cos(\omega t)sin(\omega t)-mgA \omega sin(\omega t)##

which simplifies to

##mA\omega^3 cos(\omega t) = kAcos(\omega t)-mg\omega##.

As you can see, the rate of change is driven by ##cos(\omega t)## for both.

This was quick and dirty, so if I made any errors somewhere, someone feel free to correct me.

Potential energy in a mass spring system is the energy stored in the system due to the position of the mass and the spring. It is the energy that the system has the potential to convert into other forms, such as kinetic energy.

Potential energy in a mass spring system can be calculated using the equation PE = 1/2kx^2, where PE is potential energy, k is the spring constant, and x is the displacement of the mass from its equilibrium position.

The potential energy in a mass spring system is affected by the spring constant, the displacement of the mass, and the acceleration due to gravity. It can also be affected by external forces acting on the system, such as friction or air resistance.

The potential energy in a mass spring system changes as the position of the mass changes. When the mass is at its equilibrium position, the potential energy is at its minimum. As the mass is displaced from its equilibrium position, the potential energy increases. When the mass is released, the potential energy is converted into kinetic energy as the mass oscillates back and forth.

In a mass spring system, potential energy and kinetic energy are interchangeable. As the mass moves, potential energy is converted into kinetic energy and vice versa. At the equilibrium position, all of the potential energy is converted into kinetic energy, and at the extremes of the oscillation, all of the kinetic energy is converted back into potential energy.

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