Conceptual Question About Vertical Spring System

  • #1
Suppose there is a spring with a mass m attached to it, and it is released from rest. It will oscillate, starting from a certain point y0, going back to equilibrium and then down to let's say y2. Let's say in the first case, I call y1 my x = 0 point for the spring. When it passes equilibrium, is it not true that mg = kx since the acceleration is zero? However, let's repeat the case for which i will call the equilibrium point the x=0 for the spring. Now, from energy we have: mgx = kx^2/2. This means that 2mg = kx. How can this be possible?
 

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  • #2
PhanthomJay
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Suppose there is a spring with a mass m attached to it, and it is released from rest. It will oscillate, starting from a certain point y0, going back to equilibrium and then down to let's say y2. Let's say in the first case, I call y1 my x = 0 point for the spring. When it passes equilibrium, is it not true that mg = kx since the acceleration is zero?
yes
However, let's repeat the case for which i will call the equilibrium point the x=0 for the spring. Now, from energy we have: mgx = kx^2/2. This means that 2mg = kx. How can this be possible?
You are mixing apples and oranges. For the 2nd case, the mass has kinetic energy at x= 0 (point y1). You are better off using the energy equations from the starting point y0. Then from the 2 'no speed' positions (at release and at the end point when the mass comes to a temporary stop), then mg(2x) = 1/2(k(2x)^2 , k = mg/x. Note that the stretch is double the stretch if the mass was slowly lowered to its final at rest position y1.
 
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  • #3
Thanks for your help; that's brilliant. However, wouldn't the stretch be the same even if it was lowered? For the first equation: x = mg/k. For the one slowly lowered, since the forces are balanced once it reaches equilibrium, shouldn't the stretch be x = mg/k also?
 
  • #4
PhanthomJay
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However, wouldn't the stretch be the same even if it was lowered? For the first equation: x = mg/k. For the one slowly lowered, since the forces are balanced once it reaches equilibrium, shouldn't the stretch be x = mg/k also?
Yes. When it is slowly lowered with your hand applying an external force, then you release it once it reaches it's equilibrium position, the stretch is mg/k. Or, if it is damped due to friction, it will ultimately, after oscillating back and forth a few times, settle to that same point, x =mg/k.
 

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