Springs and shm

  • Thread starter Gregg
  • Start date
  • #1
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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 isnt it included in the potential and kinetic energy?
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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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
 
  • #3
459
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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?
 
  • #4
rock.freak667
Homework Helper
6,230
31
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.
 

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