Verifying Conservation of Energy for Vertical Spring Oscillator

[Soaring Crane]

Show that the conservation of energy holds also for the vertical spring where x is measured from the vertical equilibrium position (x_0 = mg/k) for a mass m.

This is what I did so far. Where do I go from here?

f = mg - kx_0 = 0
kx_0 = mg
E = KE + PE_grav + PE_spring
= ([mv^2]/2) - mg(x + x_0) + [(x + x_0)^2]/2
= ([mv^2]/2) - mgx - mgx_0 + (k/2)[x^2 + 2xx_0 + x_0^2]
= ([mv^2]/2) - mgx - mgx_0 + [(kx^2)/2] + xx_0k + [(kx_0^2)/(2)]
E_x0 = ([mv^2]/2) - mgx_0 + [(kx_0^2)/2]

Thanks for helping.

 

[Integral]

Consider finding an expression for your velocity in terms of the position. Then compare energy at the equilibrium postion to that at maximum extension.

Also please consider reading through this thread to learn how to present equations in a more readable format.

 

[wais]

To continue, you can substitute mg/k for x_0 in the equation for E_x0 to get:

E_x0 = ([mv^2]/2) - mg(mg/k) + [(k(mg/k)^2)/2]
= ([mv^2]/2) - (mg^2)/k + (mg^2)/(2k)
= ([mv^2]/2) - (mg^2)/(2k) + (mg^2)/(2k)
= ([mv^2]/2)

Since this is the same as the equation for E, we can conclude that E_x0 = E, showing that the conservation of energy holds for the vertical spring oscillator. This means that the total energy (kinetic energy + potential energy) remains constant throughout the oscillation, regardless of the position of the mass. This is a fundamental principle of physics and is a crucial concept in understanding the behavior of systems such as the vertical spring oscillator.