Springs and shm

1. May 18, 2010

Gregg

Need to show that the vertical displacement of a particle on a spring is

$$\ddot{x} + 100x = 0$$

$$\frac{1}{2}kx^2 + \frac{1}{2}m\dot{x}^2=C$$

$$k\dot{x}x + m\dot{x}\ddot{x}=0$$

Then since m=0.4, k=40.

$$\ddot{x}+100x=0$$

what has happened to gravitational potential energy? why isnt it included in the potential and kinetic energy?

2. May 18, 2010

rock.freak667

Well you see if you include it, it will cancel out due to equilibrium conditions

$$\frac{1}{2}k(x+\delta)^2 -mgx + \frac{1}{2}m\dot{x}^2 = C$$

At equilibrium kδ=mg or kδ-mg = 0

3. May 18, 2010

Gregg

ah i see how it vanishes now. but does
$$\frac{1}{2}k\delta ^2$$
mean anything? does it disappear?

it does because its a constant right?

4. May 18, 2010

rock.freak667

Remember that when you have a spring and then you suspend a mass from it, there will be an initial displacement. δ is this displacement such that when you displace the mass a distance 'x', the spring extends by 'δ+x'

Also it disappears after you differentiate the energy equation.