Work done pushing a spring from the side

[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: $$ W_{me} = 2k \int_a^b \frac {y \sqrt {y^2 - l}} {\frac {l} {2} + \sqrt {y^2 - l}} \, dy $$ $$ = 2k \int_a^b \frac {y} {\frac {l} {2 \sqrt {y^2 - l}} + 1} \, dy $$

 

[kuruman]

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.

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?

 

[erobz]

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?

 

[Steve4Physics]

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.

 

[Ebby]

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?

 

[erobz]

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:

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

 

[Ebby]

@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]

@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.

 

[erobz]

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.

 

[Ebby]

It does work out, and it is quite efficient in comparison.

Yep I did it. Much more efficient.