Harmonic motion of a string - Find energy

[UrbanXrisis]

When a mass m, hanging from a string with spring constant k, is set into up-and-down simple harmonic motion, it has a period of vibration T, which is given by the equation T=2*pi*sqrt(m/k). The amount of elastic potential energy PE stored in the spring at any given instant is dependent on its spring constant k ant its elongation x. Determine the potential energy stored in the spring, PE, in terms of m, T, and x.

I got PE=mgh
m would be the mass in the string and h would be the elongation, but what about acceleration?

 

[Texhno]

Use the equation T=2*pi*sqrt(m/k) and solve for k. Get that and sub in what you got for k into PE=.5kx^2

PE=(4*pi^2*m*x^2)/(2*T^2)

 

[Chen]

They don't give you a relation point for the gravitational potential energy, so I will assume it to be zero at x = 0. The total potential energy is the sum of the gravitational potential energy and the elastic potential energy:

[tex]E_p = E_{p_g} + E_{p_{ele}} = mgx + \frac{1}{2}kx^2[/tex]

And K you can find from T.

 

[baffledMatt]

They don't give you a relation point for the gravitational potential energy, so I will assume it to be zero at x = 0. The total potential energy is the sum of the gravitational potential energy and the elastic potential energy:

[tex]E_p = E_{p_g} + E_{p_{ele}} = mgx + \frac{1}{2}kx^2[/tex]

And K you can find from T.

Except of course you meant
[tex]E_p = E_{p_g} + E_{p_{ele}} = -mgx + \frac{1}{2}kx^2[/tex]

 

[Chen]

Except of course you meant
[tex]E_p = E_{p_g} + E_{p_{ele}} = -mgx + \frac{1}{2}kx^2[/tex]

Of course... I didn't notice x represented elongation, I just assumed the X axis pointed up.