Study of SHM of an object attached to a vertical spring

[pgirl1729]

When applying the law of conservation of energy, to the bottom most point of the oscillation, there's only maximum elastic potential energy if we assume it to have 0 potential energy. At the center of oscillation, there's maximum kinetic energy and zero elastic potential energy. That's how my book say. Why don't you add the gravitational potential energy too?

 

[Drakkith]

I suspect the book is simplifying things so that you can focus only on the relevant content.

 

[Herman Trivilino]

When applying the law of conservation of energy, to the bottom most point of the oscillation, there's only maximum elastic potential energy if we assume it to have 0 potential energy.

Okay. YOU can make this assumption about the gravitational potential energy, but is the same assertion made in your textbook?

At the center of oscillation, there's maximum kinetic energy and zero elastic potential energy.

If that assertion about the gravitational potential energy at the lowest point has been made, then the elastic potential energy is not zero at the center of oscillation. Because the spring is under tension. The spring is not relaxed because the object hanging from the spring is stretching the spring.

That's how my book say. Why don't you add the gravitational potential energy too?

You can, and I'll explain why it's not necessary, but please first answer the question I asked above.

 

[pgirl1729]

is the same assertion made in your textbook?

No

 

[Herman Trivilino]

Okay, sorry. Neglect that bit I stated about assuming the gravitational potential energy is zero. It's not relevant. You can assume any value you want because only changes in potential energy matter.

There's a good explanation here for why the force of gravity cancels out when the object of mass $m$ is attached to the spring and stretches it to a new equilibrium height. And since the force of gravity is not relevant the potential energy associated with that force is also not relevant.

 

[john1954]

When applying the law of conservation of energy, to the bottom most point of the oscillation, there's only maximum elastic potential energy if we assume it to have 0 potential energy. At the center of oscillation, there's maximum kinetic energy and zero elastic potential energy. That's how my book say. Why don't you add the gravitational potential energy too?

Constant forces , such as gravity in this case ,can be subtracted out of the equations of motion if displacements are measured from the position of equilibrium. Write the equation of motion with gravity included and displacements measured from an arbitrary point. Then write the static equation by setting x dot dot to zero. Subtract the static equation from the dynamic equation and constants such as gravity and unstretched length will be subtracted out.

 

[john1954]

Equation of motion is

mX dot dot =mg - k(X - X_u)

Equilibrium equation is

0=mg - k(X_eq - X_u)

subtracting

mX dot dot = - k(X - X_u)+ k(X_eq - X_u)

mX dot dot = - k(X - X_eq)

and because X_eq is a constant

m(X - X_eq) dot dot = - k(X - X_eq)

Defining x= X-X_eq gives

mx dot dot=-kx

which looks like gravity is ignored - but gravity has not been ignored

 

[sophiecentaur]

At the center of oscillation, there's maximum kinetic energy and zero elastic potential energy.

Calling the PE 'elastic' was misleading (if indeed this was the wording of the book). The equilibrium position is where the net potential energy is zero - which includes both gravitational and elastic. Potential Energy is always relative and, of course, the elastic energy is not zero at equilibrium because the spring is partially stretched there. So they choose the equilibrium position for calculations because it' convenient.