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

[tex] \ddot{x} + 100x = 0 [/tex]

[tex] \frac{1}{2}kx^2 + \frac{1}{2}m\dot{x}^2=C [/tex]

[tex] k\dot{x}x + m\dot{x}\ddot{x}=0[/tex]


Then since m=0.4, k=40.

[tex] \ddot{x}+100x=0 [/tex]

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

[tex]\frac{1}{2}k(x+\delta)^2 -mgx + \frac{1}{2}m\dot{x}^2 = C[/tex]


At equilibrium kδ=mg or kδ-mg = 0

 

[Gregg]

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

[tex]\frac{1}{2}k(x+\delta)^2 -mgx + \frac{1}{2}m\dot{x}^2 = C[/tex]


At equilibrium kδ=mg or kδ-mg = 0

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


it does because its a constant right?

 

[rock.freak667]

ah i see how it vanishes now. but does
[tex] \frac{1}{2}k\delta ^2 [/tex]
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.

 

[rdl]

The equations provided show the relationship between the vertical displacement (x) of a particle on a spring and the forces acting on it, specifically the spring constant (k) and the mass (m). The first equation, \ddot{x} + 100x = 0, is the equation of motion for the particle, where the acceleration (\ddot{x}) is equal to the negative of the displacement (x) multiplied by the spring constant (100). This equation shows that the particle will oscillate up and down on the spring with a period determined by the spring constant and the mass.

The second equation, \frac{1}{2}kx^2 + \frac{1}{2}m\dot{x}^2=C, represents the conservation of energy for the particle. The left side of the equation represents the total energy of the system, which is the sum of the potential energy stored in the spring (\frac{1}{2}kx^2) and the kinetic energy of the particle (\frac{1}{2}m\dot{x}^2). The right side of the equation, represented by the constant C, shows that the total energy remains constant throughout the motion.

The third equation, k\dot{x}x + m\dot{x}\ddot{x}=0, is a useful relationship that can be derived from the first two equations. It shows that the work done by the forces (kx and m\ddot{x}) is equal to zero, meaning that the forces are conservative. This is a result of the potential energy being solely dependent on the displacement and not on the velocity of the particle.

As for the question about gravitational potential energy, it is not included in these equations because they are specifically for a particle on a spring. In this system, the only forces acting on the particle are the spring force and the force of gravity is negligible. If we were to consider a more complex system where the particle is affected by gravity, then the equations would need to be modified to include gravitational potential energy. However, in this case, it is not necessary to include it as it does not significantly affect the motion of the particle on the spring.