Conceptual Question About Vertical Spring System

[lvslugger36]

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?

 

[PhanthomJay]

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.

 

[lvslugger36]

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?

 

[PhanthomJay]

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.

 

[rwisz]

It is important to note that the equations mg = kx and mgx = kx^2/2 represent two different scenarios and cannot be compared in this way. In the first case, we are considering the forces acting on the mass at a specific point in time, when it is passing through equilibrium. At this point, the acceleration is zero and the forces are balanced, so mg = kx. However, in the second case, we are looking at the total energy of the system, which includes the potential energy stored in the spring. This energy is dependent on the displacement of the mass from the equilibrium point, hence the equation mgx = kx^2/2. It is not valid to equate these two equations as they represent different aspects of the system. The value of kx in the second equation represents the maximum displacement of the mass, not its position at a specific point in time. Therefore, it is possible for 2mg to equal kx in the second case, as it is a different scenario with different variables.