Work done pushing a spring from the side

• Ebby
In summary, the conversation discusses the formation of an integral for the work done and the use of two identical springs with a spring constant of 2k. The resulting force and work equations are corrected, and the integral formula is adjusted to account for the use of two springs.
Ebby
Homework Statement
How much work must you do to push the midpoint of the string up or down a distance y?
Relevant Equations
F = -kx
W = Fy

My question is whether I've formed the integral for the work done correctly? It just seems a bit unwieldy to me...

If I call the extension of the spring ## x ##, I can see that ## z = \frac l 2 + x ## and ## z^2 = \left( \frac {l} {2} \right)^2 + y^2 ##. Combining them gives: $$x = \sqrt {y^2 - l}$$

Since the restoring force generated by one spring is ## F_{res} = -k \sqrt {y^2 - l} ## along its axis, the force that must be exerted by me to overcome both springs is: $$F_{me} = 2k \sqrt {y^2 - l}$$

Now, using ## W = \int |\vec F| \, \cos \theta \, |d \vec r| ## where ## cos \theta = \frac {y} {z} = \frac {y} {\frac {l} {2} + x} ## we can say that: $$W_{me} = 2k \int_a^b \frac {y \sqrt {y^2 - l}} {\frac {l} {2} + \sqrt {y^2 - l}} \, dy$$ $$= 2k \int_a^b \frac {y} {\frac {l} {2 \sqrt {y^2 - l}} + 1} \, dy$$

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First of all your equation ##x = \sqrt {y^2 - l}## is dimensionally incorrect and that error propagates down your derivation. I'm sure that's a typo.
Ebby said:
Since the restoring force generated by one spring is ## F_{res} = -k \sqrt {y^2 - l} ## along its axis, the force that must be exerted by me to overcome both springs is: $$F_{me} = 2k \sqrt {y^2 - l}$$
Why the doubling? Is this how we add vectors that have the same magnitude but point in different directions?

Don't we have to be careful about the spring constant? Imagine two springs of length ##l/2## connected in series. If the full spring length ##l## has a spring constant of ##k##, then each spring of length ##l/2## must have spring constant ##k_{l/2} = ?##

Also…maybe the math of the dot product takes care of it, but I get something less messy by noticing something about the "work" from the horizontal components of the restoring force through the displacement?

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Ebby said:
My question is whether I've formed the integral for the work done correctly? It just seems a bit unwieldy to me...
If you are allowed to use the standard formula for the elastic potential energy of a spring ##(E = \frac 12 kx^2)##, then part (a) can be done without calculus. Simply consider the total extension.

malawi_glenn and Lnewqban
Yeah I really messed that up. Take two.

I shall replace the single spring with two identical springs, each with a spring constant of ## 2k ##. This is equivalent.

Considering one of the springs, if I call its extension ## x ##, I can see that ## z = \frac l 2 + x ## and ## z^2 = \left( \frac {l} {2} \right)^2 + y^2 ##.

Combining gives: $$x^2 + lx - y^2 = 0$$ Solving for x: $$x = \frac {-l \pm \sqrt {l^2 + 4y^2}} {2}$$ I only want the positive solution, so it's: $$x = \frac {-l + \sqrt {l^2 + 4y^2}} {2}$$ Now, the force with which I push against each spring is: $$F_{each} = 2kx = k(-l + \sqrt {l^2 + 4y^2})$$ The force in the ## y ## direction is: $$F_{each_{y}} = k(-l + \sqrt {l^2 + 4y^2})\cos\theta$$ $$= \frac {k(-l + \sqrt {l^2 + 4y^2})y} {\frac l 2 + \frac {-l + \sqrt {l^2 + 4y^2}} {2}}$$ $$= \frac {2k(-l + \sqrt {l^2 + 4y^2})y} {\sqrt {l^2 + 4y^2}}$$ $$= 2ky(\frac {-l} {\sqrt {l^2 + 4y^2}} +1)$$ So the work done pushing against one spring is [EDIT: inserted missing integration symbol]: $$W_{each_{y}} = 2k \int_a^b (\frac {-l} {\sqrt {l^2 + 4y^2}} +1)y \, dy$$ And against two springs [EDIT: inserted missing integration symbol]: $$W_{both_{y}} = 4k \int_a^b (\frac {-l} {\sqrt {l^2 + 4y^2}} +1)y \, dy$$ Does this integral look right now?

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Ebby said:
Yeah I really messed that up. Take two.

I shall replace the single spring with two identical springs, each with a spring constant of ## 2k ##. This is equivalent.

Considering one of the springs, if I call its extension ## x ##, I can see that ## z = \frac l 2 + x ## and ## z^2 = \left( \frac {l} {2} \right)^2 + y^2 ##.

Combining gives: $$x^2 + lx - y^2 = 0$$ Solving for x: $$x = \frac {-l \pm \sqrt {l^2 + 4y^2}} {2}$$ I only want the positive solution, so it's: $$x = \frac {-l + \sqrt {l^2 + 4y^2}} {2}$$ Now, the force with which I push against each spring is: $$F_{each} = 2kx = k(-l + \sqrt {l^2 + 4y^2})$$ The force in the ## y ## direction is: $$F_{each_{y}} = k(-l + \sqrt {l^2 + 4y^2})\cos\theta$$ $$= \frac {k(-l + \sqrt {l^2 + 4y^2})y} {\frac l 2 + \frac {-l + \sqrt {l^2 + 4y^2}} {2}}$$ $$= \frac {2k(-l + \sqrt {l^2 + 4y^2})y} {\sqrt {l^2 + 4y^2}}$$ $$= 2ky(\frac {-l} {\sqrt {l^2 + 4y^2}} +1)$$ So the work done pushing against one spring is $$W_{each_{y}} = 2ky(\frac {-l} {\sqrt {l^2 + 4y^2}} +1) \, dy$$ And against two springs: $$W_{both_{y}} = 4ky(\frac {-l} {\sqrt {l^2 + 4y^2}} +1) \, dy$$ Does this integral look right now?
Thats what I got, but the extra algebra trying to use the extension can be avoided:

$$F_{l/2} = k_{l/2} \left( z - \frac{l}{2}\right)$$

The vertical component of each spring is

$$F_{y_{l/2}} = k_{l/2} \left( z - \frac{l}{2}\right) \cos \theta = k_{l/2} \left( z - \frac{l}{2}\right) \frac{y}{z} = k_{l/2} \left( 1 - \frac{l}{2 z}\right) y$$

Then multiply by 2 for each spring contribution:

$$dW = F_{y_{l/2}}~dy = 2 k_{l/2} \left( 1 - \frac{l}{2 \sqrt{ (l/2)^2 + y^2}}\right) y~dy$$

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@erobz Yes that's better. (Btw, I know you meant ## \cos\theta ##.)

I'm just going over this to see if I can get the same answer from the elastic potential energy equation suggested by @Steve4Physics.

erobz
Ebby said:
@erobz Yes that's better. (Btw, I know you meant ## \cos\theta ##.)

I'm just going over this to see if I can get the same answer from the elastic potential energy equation suggested by @Steve4Physics.
Oh, I chose the opposite angle in my working. Didn’t notice you hadn’t picked that one in your diagram. I’ll fix it.

Ebby said:
I'm just going over this to see if I can get the same answer from the elastic potential energy equation suggested by @Steve4Physics.
It does work out, and it is quite efficient in comparison.

Steve4Physics and Lnewqban
erobz said:
It does work out, and it is quite efficient in comparison.
Yep I did it. Much more efficient.

erobz and Steve4Physics

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