Spring is stretched, then attached to a mass, what is its speed at x?

  • #1
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Homework Statement:
A hanging spring (k=100) is in equilibrium (without the mass M). It is then pulled down to 50 cm from equilibrium. Now a mass of 3 kg is attached and the spring released.

a) what is the speed of the mass after it has moved up 10 cm?
b) what are the heights at which the mass has a speed of 0.8 m/s?
Relevant Equations:
ma = -kx
E = mgx + 1/2kx^2 + 1/2mv^2
Dear all,

I am back with another Spring problem.
I have tried to use the insights I gained from your help last time:
https://www.physicsforums.com/threa...nded-from-a-spring.972942/page-3#post-6190934

I figured I start with calculating the new equilibrium by ## x =\frac{-mg}{k} ##
And then subtracting that from the 50 cm to get the amplitude which gave me -20.6 cm.

I then set up the energy equation with a fixed coordinate system as I learned in the above thread:
With the -20.6 being A.

## E = \frac {1}{2}*m*v^2+\frac {1}{2}*k*x^2+mgx ##
so
## \frac {1}{2}*k*A^2+mgA = \frac {1}{2}*m*v^2+\frac {1}{2}*k*x^2+mgx ##

My solution and calculation are in the pictures attached.

For b) I could only think of using the quadratic formula, is there a better way?

Is my work correct?

Thank you all
 

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Answers and Replies

  • #2
kuruman
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The amplitude is always positive so it should be A = 20.6 cm = 0.206 m. Reread my post #54 in the previous thread. The better way is to take as the zero of the potential energy (gravitational and elastic) the new equilibrium position which you have already calculated. Then the total mechanical energy with respect to that zero is ##ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2## where ##x## is the displacement from the new equilibrium position.

Note that the equation for the ##ME## is a quadratic, but simpler than the one you contemplate. You do need to solve a quadratic because there two heights at which the speed has the same value, one is below the new equilibrium position and the other is symmetrically above it.
 
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  • #3
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The amplitude is always positive so it should be A = 20.6 cm = 0.206 m. Reread my post #54 in the previous thread. The better way is to take as the zero of the potential energy (gravitational and elastic) the new equilibrium position which you have already calculated. Then the total mechanical energy with respect to that zero is ##ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2## where ##x## is the displacement from the new equilibrium position.

Note that the equation for the ##ME## is a quadratic, but simpler than the one you contemplate. You do need to solve a quadratic because there two heights at which the speed has the same value, one is below the new equilibrium position and the other is symmetrically above it.

Hi Kuruman,

thank you. I reread your post.

I thought that I had used the equilibrium position as zero for the gravitational potential energy?
 
  • #4
PeroK
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Hi Kuruman,

thank you. I reread your post.

I thought that I had used the equilibrium position as zero for the gravitational potential energy?
Is the equilibrium position the same with or without the hanging mass?
 
  • #5
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Is the equilibrium position the same with or without the hanging mass?

I think it is not, so I adjusted for that. Otherwise, A would be 0.50 m no? Instead of 0.206 and x of mgx would be -0.50 instead of -0.206
I'm sorry if I'm somewhat slow to catch on.
 
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  • #6
kuruman
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OK, I will try to explain it differently and repeat myself to drive the point home.
Case I: Horizontal spring-mass system on frictionless surface oscillating with amplitude ##A##
$$ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$$
Case II: Vertical spring-mass system oscillating with amplitude ##A##
$$ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$$
Note: In case II ##x## is the distance from the equilibrium position, that is the level at which the mass is hanging at rest when it is not oscillating. When you measure displacement relative to this equilibrium position you do not have to worry about gravitational potential energy, only spring elastic energy as the mass oscillates up and down.

Enrichment note: Whether the spring-mass system is vertical or horizontal or at angle ##\theta## on a frictionless incline, the mechanical energy is given by
$$ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$$
where ##A## is the amplitude of oscillations (maximum displacement from the equilibrium position) and ##x## is the instantaneous displacement from the equilibrium position ##-A \leq x \leq +A##. The equilibrium position is the position of the mass when the net force acting on it is zero.
 
  • #7
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OK, I will try to explain it differently and repeat myself to drive the point home.
Case I: Horizontal spring-mass system on frictionless surface oscillating with amplitude ##A##
$$ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$$
Case II: Vertical spring-mass system oscillating with amplitude ##A##
$$ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$$
Note: In case II ##x## is the distance from the equilibrium position, that is the level at which the mass is hanging at rest when it is not oscillating. When you measure displacement relative to this equilibrium position you do not have to worry about gravitational potential energy, only spring elastic energy as the mass oscillates up and down.

Enrichment note: Whether the spring-mass system is vertical or horizontal or at angle ##\theta## on a frictionless incline, the mechanical energy is given by
$$ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2$$
where ##A## is the amplitude of oscillations (maximum displacement from the equilibrium position) and ##x## is the instantaneous displacement from the equilibrium position ##-A \leq x \leq +A##. The equilibrium position is the position of the mass when the net force acting on it is zero.
Thank you, Kuruman.

This is very helpful, and I think I did have this understanding but I may misunderstand the big picture.
I understand why ##ME=\frac{1}{2}kA^2=\frac{1}{2}kx^2+\frac{1}{2}mv^2##.

But, I'm probably asking a really dumb question that is already explained with your reasoning, but does the Potential Energy not influence the speed?
 
  • #8
kuruman
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But, I'm probably asking a really dumb question that is already explained with your reasoning, but does the Potential Energy not influence the speed?
Potential energy by itself does not influence the speed. The ME conservation equation may be written as ##\Delta K+\Delta U=0##. It's the changes in the two quantities that matter, not the values themselves because the zero of potential and kinetic energy is arbitrary. If you choose the unstretched position of the vertical spring as your zero of potential energy for spring and gravity, the overall change in potential energy is ##\Delta U=\Delta U_{spring}+\Delta U_{grav.}## If you choose the equilibrium position of the vertical spring as your zero of potential energy for spring and gravity, the overall change in potential energy is ##\Delta U=\Delta U'_{spring}##. The prime indicates that ##x## is measured from the equilibium position not the unstretched position of the spring. In either case the overall change ##\Delta U## is the same and equal to ##-\Delta K##.
 
  • #9
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Potential energy by itself does not influence the speed. The ME conservation equation may be written as ##\Delta K+\Delta U=0##. It's the changes in the two quantities that matter, not the values themselves because the zero of potential and kinetic energy is arbitrary. If you choose the unstretched position of the vertical spring as your zero of potential energy for spring and gravity, the overall change in potential energy is ##\Delta U=\Delta U_{spring}+\Delta U_{grav.}## If you choose the equilibrium position of the vertical spring as your zero of potential energy for spring and gravity, the overall change in potential energy is ##\Delta U=\Delta U'_{spring}##. The prime indicates that ##x## is measured from the equilibium position not the unstretched position of the spring. In either case the overall change ##\Delta U## is the same and equal to ##-\Delta K##.
Hi Kuruman,

thank you. I have trouble. I'm sorry. I drew it just now, but...

I've worked with your reasoning, up until ##\Delta U=\Delta U'_{spring}##
Once Mass and spring move from said zero (equilibrium), why do we not account for the change in potential gravitational energy? Please believe me that I'm trying really hard to understand.

I meant isn't some of the Potential Gravitational Energy converted to Kinetic energy and back?
 
  • #10
kuruman
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I meant isn't some of the Potential Gravitational Energy converted to Kinetic energy and back?
It is but by proper choice of the zero of gravitational potential energy and spring energy, you don't have to worry about the change in the gravitational part of the change in potential energy. That's the bottom line of post #54 in the other thread. Let me say the same thing a bit differently.
When both zeroes of potential energy are where the mass is at the unstreched position of the spring the potential energy functions are
$$U_g=m g x~;~~~U_s=\frac{1}{2}kx^2$$
Since both potential energies are zero when ##x=0##, the total change in potential energy when the mass goes from the unstretched position to ##x## is $$\Delta U_{total}=\Delta U_g+\Delta U_s=U_g+U_s=m g x+\frac{1}{2}kx^2$$
Now suppose I choose the zeroes of potential energies to be at the equilibrium position of the spring, ##x_{eq}=mg/k##. Furthermore, let me pick the origin of coordinates also there. Let ##\xi## be a parameter measuring the displacement of the mass from that position. Then the gravitational potential energy is $$U_g=mg\xi.$$ The spring energy is a bit trickier. I need to re-express ##x## in terms of my new parameter ##\xi##. The correct expression is ##x=\xi-x_{eq.}##. It says that when ##\xi## is a negative stretch below the equilibrium position, the magnitude of ##x## is larger and the opposite when the spring is compressed from the equilibrium position when ##\xi## is positive. I also need to subtract ##\frac{1}{2}kx_{eq}^2## from the spring potential energy in order to move its zero to the equilibrium point. Thus, the original ##U_s## is rewritten as follows
$$U_s=\frac{1}{2}kx^2~\rightarrow ~\frac{1}{2}k(\xi-x_{eq})^2-\frac{1}{2}kx_{eq}^2=\frac{1}{2}k\xi^2-k\xi x_{eq}.$$
As before, the total change in potential energy when the mass is displaced by ##\xi## from the equilibrium position (where both potential energies are zero) is
$$\Delta U_{total}=U_g+U_s=mg\xi + \frac{1}{2}k\xi^2-k\xi x_{eq}=mg\xi +\frac{1}{2}k\xi^2-k\xi \frac{mg}{k}=\frac{1}{2}k\xi^2$$
As you can see the change in gravitational potential energy was not ignored. By choosing zeroes of potential energy the way I did, the linear cross-term in the change in spring potential energy exactly cancels the change in gravitational potential energy. What's left in the total potential energy change is spring energy. Furthermore, this left over spring energy is as if the spring were initially unstretched at the equilibrium position and then stretched (or compressed) by additional amount ##\xi##.

I hope this clarifies everything for you.
 
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  • #11
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Hi Kuruman,

Thank you. Seeing that the term cancels has helped.
I will try to rework the problem to let it sink in and come back.
 

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