Consider the case of a mass oscillating up and down on a spring. If you measure

Question

Consider the case of a mass oscillating up and down on a spring. If you measure your distances from the unstretched (x = 0) position of the spring, the total energy of the system is: Conservation of Energy - 8 E = 1/2mv^2 + 1/2kx^2 - mgx (Equation 1) However, we can make our lives much easier in Part II of the lab if we measure from the equilibrium (x = - mg/k) position of the spring because the gravitational potential energy (-mgx) appears to vanish: E = 1/2mv^2 + 1/2kx^2_EQ (Equation 2) Make the substitution x_EQ = x - mg/k into Equation 2 and show that the gravitational potential energy, reappears. You will also get a constant term. Why can you ignore it?

Explanation / Answer

E = 1/2mv^2+1/2k(x-mg/k)^2

E = 1/2mv^2+1/2k(x^2+m^2g^2/k^2-2mgx/k)

E = 1/2mv^2+1/2kx^2+m^2g^2/2k-mgx

here mgx is the gravitaional energy term

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