How Does Adding a Second Mass Affect the Dynamics of a Vertical Spring?

In summary: The diagram just defines h as a measure of the distance between the mass and its equilibrium position.
  • #1
Thefox14
40
0

Homework Statement



[PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-08-GPE-ME/twomasses_vertical_spring/3.gif

A block of mass 7 kg hangs on a spring. When a second block with an identical mass of 7 kg is tied to the first, the spring stretches an additional ho = 1.3 m.

a) What is the value of the spring constant k?

Now the string is burned and the second block falls off.
b) How far above its original position does the remaining block attain its maximum speed?

c) What is the maximum speed attained by the remaining block?

Homework Equations



[tex]W_{Total} = \Delta K[/tex]
[tex]W_{NC} = \Delta ME[/tex]

The Attempt at a Solution



I was able to figure out part a and b fairly easily.
A) was 52.82 N/m
B) was 1.3 m

Part c is where I'm stuck. So far what I've done is set the point where the string is at equilibrium with no weight on it where potential energy = 0. Since there are no Non Conservative forces I sad:

[tex]\frac{1}{2}k(2h)^{2} - mg(2h) = \frac{1}{2}mv_{f}^{2}[/tex]
Where 2h is twice the answer found in B
I get the left side to be a negative number, what am i doing wrong?
 
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  • #2
Thefox14 said:
Part c is where I'm stuck. So far what I've done is set the point where the string is at equilibrium with no weight on it where potential energy = 0. Since there are no Non Conservative forces I sad:

[tex]\frac{1}{2}k(2h)^{2} - mg(2h) = \frac{1}{2}mv_{f}^{2}[/tex]
Where 2h is twice the answer found in B
I get the left side to be a negative number, what am i doing wrong?
Your conservation of energy idea is a good idea. But equation isn't set up quite right.

For what I'm about to say, refer back to the figure (maybe a few times). There's more than one way to approach this problem, but as long as your figure has defined h and h0 for you already, we should use them as they are defined.

To start, find the total energy of the system when the spring is stretched to its maximum (i.e. when h = 0).

There are 3 things to consider: the spring's potential energy, the block's kinetic energy, and the gravitational potential energy. But when h = 0, the block is not moving, so the kinetic energy is zero. And we can define the reference point for gravitational potential at any height we choose. Just to make things simple, define the graviational potential energy to be zero when h = 0. So that just leaves the spring. What is the spring's potential energy when h = 0? (Hint: this will be the total energy of the system; and it is a function solely of k and h0).

As time goes on, after the string breaks, the total energy of the system remains constant. So set the total energy of the system equal to the sum of the following:
o The potential energy of the spring (a function of k, h0, and h).
o The gravitational potential energy (a function of m, g, and h).
o The kinetic energy (a function of m and v).

(Note that since we defined the gravitational potential energy to be 0 when h = 0, the gravitational potential energy is not a function of h0. Refer again to your attached figure if you need to.)

Solve for v when h = h0 (and you might want to make note that h0 = mg/k).

Good luck! :smile:.
 
  • #3
I'm not quite sure what h represents in the diagram as its pretty much the same as h0. I know h0 is the amount the spring has stretched from its equilibrium though.

It sounds like you're setting the potential energy to be 0 when the spring is stretched to its max. So U0 would be (1/2)kx2, with the x equal to the total amount stretched correct? So it would be 2.6m (2* 1.3m). The place where kinetic energy is the highest would be where the spring is relaxed without the mass on it right? So wouldn't this be h0 = 0 and (1/2)mv2 + mgh with h being 2.6m?

Could you explain what h is? I think that's the one part I'm missing.

Thanks for the reply!
 
  • #4
Thefox14 said:
I'm not quite sure what h represents in the diagram as its pretty much the same as h0. I know h0 is the amount the spring has stretched from its equilibrium though.

It sounds like you're setting the potential energy to be 0 when the spring is stretched to its max. So U0 would be (1/2)kx2, with the x equal to the total amount stretched correct? So it would be 2.6m (2* 1.3m).
Yes, that's right!

But just for clarity, when we talk of potential energy, there are two types of potential energy here. There's the spring's potential energy which is maximum when h = 0, and there is also the gravitational potential energy. We are defining the gravitational potential energy to be 0 when h = 0 (we don't have to do this -- we could assign the gravitational potential energy to be zero at any height we wish. But it makes the math easier if we do it this way). By doing it this way, when h = 0 all the energy in the system is contained in the spring.
The place where kinetic energy is the highest would be where the spring is relaxed without the mass on it right? So wouldn't this be h0 = 0 and (1/2)mv2 + mgh with h being 2.6m?
Look at the diagram again. h0 is a constant and does not vary. The variable h is what varies.

With just the one block on the spring, the maximum kinetic energy is when h = h0. (When h = 2h0, the block's kinetic energy is zero, the spring's potential energy is zero, and all the energy in the system is in the gravitational energy.)
Could you explain what h is? I think that's the one part I'm missing.
h is simply the height of the mass above its minimum amplitude. As the mass oscillates, h varies between 0 and 2h0.
 
  • #5
collinsmark said:
With just the one block on the spring, the maximum kinetic energy is when h = h0.

Didn't realize that haha, as soon as you said that i got the problem thanks!

Sorry to ask, but there is another problem very similar to this that I have applied the same principles and ideas to and can't get the right answer (I get very close to it though). Would you mind taking a quick look at it?
 
  • #6
Sure, but you should start a new thread for a new problem. Once you post the new thread, PM me and I'll try to take a look at it.
 

Related to How Does Adding a Second Mass Affect the Dynamics of a Vertical Spring?

What is the concept of "Two Masses on Vertical Spring"?

The concept of "Two Masses on Vertical Spring" is a classic physics problem that involves two objects, or masses, connected by a vertical spring. The spring allows the masses to move up and down, and the goal is to determine the equations of motion for each mass.

What are the key equations used in solving "Two Masses on Vertical Spring"?

The key equations used in solving "Two Masses on Vertical Spring" are Newton's Second Law, Hooke's Law, and the conservation of energy. Newton's Second Law states that the net force on an object is equal to its mass multiplied by its acceleration. Hooke's Law relates the force exerted by a spring to its displacement. And the conservation of energy states that the total energy in a system remains constant.

How do you determine the natural frequency of the system in "Two Masses on Vertical Spring"?

The natural frequency of the system in "Two Masses on Vertical Spring" can be determined by using the equation:
fn = 1/2π √(k/m)
where fn is the natural frequency, k is the spring constant, and m is the mass of the system. This frequency is also known as the resonance frequency, and it describes the rate at which the system will oscillate when disturbed.

What is the relationship between the masses and their amplitudes in "Two Masses on Vertical Spring"?

The relationship between the masses and their amplitudes in "Two Masses on Vertical Spring" is that they are inversely proportional. This means that the larger the mass, the smaller the amplitude, and vice versa. This relationship can be seen in the equations of motion for each mass, where the amplitude is determined by the mass and the spring constant.

How does changing the initial conditions affect the motion of the masses in "Two Masses on Vertical Spring"?

Changing the initial conditions, such as the initial displacements or velocities of the masses, can have a significant impact on the motion of the masses in "Two Masses on Vertical Spring". It can affect the amplitude, frequency, and overall behavior of the system. Therefore, it is important to carefully consider and control the initial conditions in order to accurately predict and analyze the motion of the masses.

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