How Is Potential Energy Calculated in a Linear Mass-Spring System?

[loonychune]

Just wondering if you could tell me what the potential energy of this system is...?

We have a linear mass-spring system, 3 masses all of equal mass m, 4 springs all of same length, same spring constant k...

|---o---o---o---|

Each mass is displaced by x1, x2 and x3 respectively.

I am reckoning the potential energy is,

U = $$\frac{1}{2}$$k(x$$^{2}_{1}$$ + x$$^{2}_{3}$$) + $$\frac{1}{2}$$k(x$$_{3}$$ - x$$_{2}$$)$$^{2}$$ + $$\frac{1}{2}$$k(x$$_{2}$$ - x$$_{1}$$)$$^{2}$$

The superscripts in terms 2 and 3 in brackets should in fact be subscripts (but the latex code _{} is putting in superscripts)...

My means of obtaining this value are pretty raw so i would appreciate any insight into what's really going on (if of course my U is correct)...

Thanks

 

[christianjb]

There are four distances involved in the sum. Taking L to be the length between ends

x1-0
x2-x1
x3-x2
L-x3

So, you need an L in your eqn somewhere.

 

[CompuChip]

The superscripts in terms 2 and 3 in brackets should in fact be subscripts (but the latex code _{} is putting in superscripts)...

That's because you're using the TeX tags very inefficiently, what's wrong with typing:
$$U = \frac12 k (x_1^2 + x_3^2) + \frac12(x_3 - x_2)^2 + \frac12 (x_2 - x_1)^2$$
(note the entire post only has one pair of tex tags)

 

[Rainbow Child]

Your potential energy is correct!

Assuming that $x_1,x_2,x_3$ are to the right (no harm to generality), the first spring has length $l+x_1$, so
$$U_1=\frac{1}{2}\,k\,x_1^2$$
The second spring has length $l+x_2-x_1$, so
$$U_2=\frac{1}{2}\,k\,(x_2-x_1)^2$$
The third spring has length $l+x_3-x_2$, so
$$U_3=\frac{1}{2}\,k\,(x_3-x_2)^2$$
The last spring has length $l-x_3$, so
$$U_4=\frac{1}{2}\,k\,x_3^2$$
Thus

$$U = \frac12 \,k\,(x_1^2 + x_3^2) + \frac12 \,k\,(x_3 - x_2)^2 + \frac12 \,k\,(x_2 - x_1)^2$$

 

[christianjb]

Your potential energy is correct!

Assuming that $x_1,x_2,x_3$ are to the right (no harm to generality), the first spring has length $l+x_1$, so
$$U_1=\frac{1}{2}\,k\,x_1^2$$
The second spring has length $l+x_2-x_1$, so
$$U_2=\frac{1}{2}\,k\,(x_2-x_1)^2$$
The third spring has length $l+x_3-x_2$, so
$$U_3=\frac{1}{2}\,k\,(x_3-x_2)^2$$
The last spring has length $l-x_3$, so
$$U_4=\frac{1}{2}\,k\,x_3^2$$
Thus

$$U = \frac12 \,k\,(x_1^2 + x_3^2) + \frac12 \,k\,(x_3 - x_2)^2 + \frac12 \,k\,(x_2 - x_1)^2$$

A distance l-x3 corresponds to a quadratic pe term of k/2 (l-x3)^2.

 

[Rainbow Child]

A distance l-x3 corresponds to a quadratic pe term of k/2 (l-x3)^2.

No! The potential energy is

$$U=\frac12\,k\,x^2$$

where $x$ is the extension of the spring, not it's length.

 

[christianjb]

No! The potential energy is

$$U=\frac12\,k\,x^2$$

where $x$ is the extension of the spring, not it's length.

The extension and the length are the same things in this case. I'm assuming that the equilibrium length is zero, since no extra information is given.

You can see that the length must be in the energy term somewhere, because it obviously costs energy to increase the length.

 

[Rainbow Child]

I'm assuming that the equilibrium length is zero, since no extra information is given.

What do you mean by that? Each spring has length $$l$$ before we move the masses.
If the 3rd mass is displaced $x_3$ to the right, the spring's length woulb be $$l-x_3$$, thus the spring would be compressed by $x_3$

 

[loonychune]

Yes I like the way you put that rainbow child, thanks a lot.

Will have to be more efficient with [ tex ] in future n all :)