Vertical Displacement of Particle on Spring: k,m and x in Equations

[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 isn't it included in the potential and kinetic energy?

 

[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

 

[Gregg]

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]

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.