# Springs and shm

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?

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rock.freak667
Homework Helper
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

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

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?

rock.freak667
Homework Helper
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?
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.