Mass attached on two springs between the walls

[skrat]

Homework Statement

We have two springs each $0.5 m$ long and both have $k=5N/cm$. One end of the spring is attached to the wall while the other two ends are connected in the middle. On the junction point of the two springs a mass is hanged. How does the frequency of oscillation around equilibrium state depend on the initial offset due to the mass?

Hint: There is no need to explicitly write the initial offset as function of mass. Just use the offset as independent parameter using which mass can be calculated.

Homework Equations

The Attempt at a Solution

$mg+2Fsin\varphi=ma$

$mg+2k\frac{z}{\sqrt{l^2+z^2}}(\sqrt{l^2+z^2}-l)=m\ddot{z}$

Let's say that $z=D+\varepsilon$ where $\varepsilon$ is very small.

$mg+2k\frac{D+\varepsilon }{\sqrt{l^2+(D+\varepsilon )^2}}(\sqrt{l^2+(D+\varepsilon )^2}-l)=m\ddot{\varepsilon }$

$mg+2k(D+\varepsilon)-\frac{2kl(D+\varepsilon )}{l\sqrt{1+(\frac{D+\varepsilon }{l})^2}}=m\ddot{\varepsilon }$

$mg+2k(D+\varepsilon )-2k(D+\varepsilon)[1-\frac 1 2 (\frac{D+\varepsilon }{l})^2]=m\ddot{\varepsilon }$

$mg+\frac{2k}{l^2}(D+\varepsilon )^3=m\ddot{\varepsilon }$

$mg+\frac{2kD^3}{l^2}[1-3\frac \varepsilon D]=m\ddot{\varepsilon }$

$\ddot{\varepsilon }+\frac{6kD^2}{ml^2}\varepsilon =g+\frac{2kD^3}{l^2}$

So if I am not mistaken, than $\omega \propto D$. Right?

 

[BvU]

You sure you rendered the problem correctly ? Vertical walls ? Vertical oscillation ?
Is there any reason at all you may assume that D << l for your Taylor expansion ?
What if the walls are indeed vertical and 0.8 m apart with springs length 0.5 m each ?

 

[skrat]

Vertical walls? YES.

Vertical oscillations? YES. It's says that explicitly in the problem, I just forgot to translate that part.

Good point. There is a question $b$ to this problem which I did not write because I know how to answer it. And in the question b it says that the body has $50$ grams. So to answer your question: No there is no reason to assume that $D<<l$ if you read only what is written in my original post. However if you also read part b (which you couldn't), than you can find out that the spring has quite large $k$ for so little mass and therefore I think it was safe to assume that $D<<l$.

My inability to speak English properly is the answer to the last question. The problem also states that the springs are NOT deformed before the mass is applied.

 

[Chestermiller]

Hi Skrat.

If z represents the downward displacement, then the starting equation should be (note the - sign in front of the spring term):
$$mg-2k\frac{D+\varepsilon }{\sqrt{l^2+(D+\varepsilon )^2}}(\sqrt{l^2+(D+\varepsilon )^2}-l)=m\ddot{\varepsilon }$$.
Now for the approximation:
$$\sqrt{l^2+(D+\varepsilon )^2}-l≈\frac{(D+ε)^2}{2l}$$
So
$$mg-k\frac{(D+ε)^3}{l^2}=m\ddot{\varepsilon }$$
So my term for the magnitude of the spring force is half of yours.
If I continue through the analysis, I get:

$$mg-k\frac{D^3+3D^2ε}{l^2}=m\ddot{\varepsilon }$$

Notice the + sign, compared to the - sign in your equation. Note that ε = 0 defines the equilibrium position. So,

$$mg-k\frac{D^3}{l^2}=0$$
and
$$-3k\frac{D^2ε}{l^2}=m\ddot{\varepsilon }$$
What happens if you solve the first equation for D in terms of l, and substitute this into the second equation?

Chet

 

[skrat]

Hi Skrat.

If z represents the downward displacement, then the starting equation should be (note the - sign in front of the spring term):
$$mg-2k\frac{D+\varepsilon }{\sqrt{l^2+(D+\varepsilon )^2}}(\sqrt{l^2+(D+\varepsilon )^2}-l)=m\ddot{\varepsilon }$$.
Now for the approximation:
$$\sqrt{l^2+(D+\varepsilon )^2}-l≈\frac{(D+ε)^2}{2l}$$
So
$$mg-k\frac{(D+ε)^3}{l^2}=m\ddot{\varepsilon }$$

I have absolutely no idea how you found this out. OR even less why my method didn't bring me to the same result with a sign error.

Ok, I agree with $$\sqrt{l^2+(D+\varepsilon )^2}-l≈\frac{(D+ε)^2}{2l}$$ which than leaves you with

$mg-\frac{2k(D+\varepsilon )}{\sqrt{l^2+(D+\varepsilon )^2}} \frac{(D+\varepsilon)^2}{2l}=m\ddot{z}$

Now I am assuming you tried to approximate $\frac{1}{\sqrt{l^2+(D+\varepsilon )^2}}$. I don't know how you did it but my way brings me to a completely different conclusion:

$\frac{1}{\sqrt{l^2+(D+\varepsilon )^2}}=\frac{1}{l\sqrt{1+(\frac{D+\varepsilon }{l})^2}}≈\frac 1 l (1-\frac{1}{2} (\frac{D+\varepsilon }{l})^2+...)$

Meaning:

$mg -2k(D+\varepsilon )\frac 1 l (1-\frac{1}{2} (\frac{D+\varepsilon }{l})^2)\frac{(D+\varepsilon)^2}{2l}=m\ddot{z}$

$mg-\frac{k(D+\varepsilon )^3}{l^2}+\frac{k(D+\varepsilon )^5}{4l^4}=m\ddot{z}$

Now It looks like you completely ignored the last term. Why would you do that? It also includes $ε$ just as the second term does.Now to answer your last question $-3k\frac{D^2ε}{l^2}=m\ddot{\varepsilon }$

$m\ddot{\varepsilon }+3k\frac{D^2ε}{l^2}=0$

$\ddot{\varepsilon }+3k\frac{1}{ml^2}(\frac{mgl^2}{k})^{2/3}ε=0$

right?

 

[skrat]

:)

Btw, thanks for all the help so far

 

[Chestermiller]

I have absolutely no idea how you found this out. OR even less why my method didn't bring me to the same result with a sign error.

Ok, I agree with $$\sqrt{l^2+(D+\varepsilon )^2}-l≈\frac{(D+ε)^2}{2l}$$ which than leaves you with

$mg-\frac{2k(D+\varepsilon )}{\sqrt{l^2+(D+\varepsilon )^2}} \frac{(D+\varepsilon)^2}{2l}=m\ddot{z}$

Now I am assuming you tried to approximate $\frac{1}{\sqrt{l^2+(D+\varepsilon )^2}}$. I don't know how you did it but my way brings me to a completely different conclusion:

$\frac{1}{\sqrt{l^2+(D+\varepsilon )^2}}=\frac{1}{l\sqrt{1+(\frac{D+\varepsilon }{l})^2}}≈\frac 1 l (1-\frac{1}{2} (\frac{D+\varepsilon }{l})^2+...)$

Meaning:

$mg -2k(D+\varepsilon )\frac 1 l (1-\frac{1}{2} (\frac{D+\varepsilon }{l})^2)\frac{(D+\varepsilon)^2}{2l}=m\ddot{z}$

$mg-\frac{k(D+\varepsilon )^3}{l^2}+\frac{k(D+\varepsilon )^5}{4l^4}=m\ddot{z}$

Now It looks like you completely ignored the last term. Why would you do that? It also includes $ε$ just as the second term does.

Within the framework of the approximation we've made, that term is negligible compared to the second term because $\frac{1}{2} (\frac{D+\varepsilon }{l})^2<<1$

Now to answer your last question $-3k\frac{D^2ε}{l^2}=m\ddot{\varepsilon }$

$m\ddot{\varepsilon }+3k\frac{D^2ε}{l^2}=0$

$\ddot{\varepsilon }+3k\frac{1}{ml^2}(\frac{mgl^2}{k})^{2/3}ε=0$

right?

I haven't checked your algebra, but I'm assuming it's correct. But, I think you should re-do the coefficient of ε so that the exponents on each of the parameters are combined to a single exponent.

Chet

 

[BruceW]

I've been looking at this thread, but I still have no idea what the set-up is meant to be... There are two vertical walls either side of a mass, which is held there because of two springs (each of which is attached to the mass and to one of the walls). And the idea is that the mass undergoes vertical oscillations. Did I understand it all OK? If so, I don't understand the reason for your starting equation...

 

[BruceW]

I've attached a picture of what I think the experimental set-up is. Have I understood it correctly? p.s. the two wiggly lines are meant to be springs.

 

[skrat]

I've attached a picture of what I think the experimental set-up is. Have I understood it correctly? p.s. the two wiggly lines are meant to be springs.

You have.

So, what exactly is supposed to be wrong with my starting equation?

 

[BruceW]

Generally, the force on the mass due to a spring is: $\vec{F}=-k\vec{d}$ where $\vec{d}$ is the displacement vector pointing from the equilibrium position of the spring, to the mass itself. Or in this case, you can just think of $-\vec{d}$ as having a direction which points from the mass, towards the place where the spring attaches to the wall. So now if you have two springs attached to opposite walls, what do you get by adding the two forces together? (If you think about it this way, it really seems like your starting equation does not make sense. So I think one of us has not got the right starting equation).

 

[skrat]

Generally, the force on the mass due to a spring is: $\vec{F}=-k\vec{d}$ where $\vec{d}$ is the displacement vector pointing from the equilibrium position of the spring, to the mass itself. Or in this case, you can just think of $-\vec{d}$ as having a direction which points from the mass, towards the place where the spring attaches to the wall. So now if you have two springs attached to opposite walls, what do you get by adding the two forces together?

Twice as much force on the same mass? But I have factor 2 in my starting equation. I guess you're trying to warn me about the sign error in my post #1. If that's the case, Chet already mentioned that in post #4.

Or maybe I completely misunderstood what you are trying to say?

 

[ehild]

Generally, the force on the mass due to a spring is: $\vec{F}=-k\vec{d}$ where $\vec{d}$ is the displacement vector pointing from the equilibrium position of the spring, to the mass itself. Or in this case, you can just think of $-\vec{d}$ as having a direction which points from the mass, towards the place where the spring attaches to the wall. So now if you have two springs attached to opposite walls, what do you get by adding the two forces together? (If you think about it this way, it really seems like your starting equation does not make sense. So I think one of us has not got the right starting equation).

Show your starting equation.


ehild

 

[BruceW]

hmm. OK, let's say $z$ is in the 'down' direction (similar to how skrat has defined it). And say $l$ is half the distance between the two walls. So the force due to the left spring is: $-k(z\hat{z}+l\hat{x})$ (where $\hat{x}$ is the horizontal direction). And the force due to the right spring is: $-k(z\hat{z}-l\hat{x})$. And the force due to gravity is simply $mg\hat{z}$ with $g$ being 9.8m/s^2. So the resultant force is:
$$\vec{F} = -k(z\hat{z}+l\hat{x}) -k(z\hat{z}-l\hat{x}) + mg\hat{z}$$
And now, cancelling some terms, and assuming that the only acceleration is in the z direction, we get:
$$m\ddot{z} = -2kz + mg$$

edit: no wait, $z\hat{z}+l\hat{x}$ is the displacement from where the left spring attaches to the wall. So the displacement from the equilibrium position of the spring would instead be
$$z\hat{z}+l\hat{x}-l\frac{z\hat{z}+l\hat{x}}{\sqrt{z^2+l^2}}$$
And adding this to the force due to the other spring gives:
$$-2kz\hat{z}(1-\frac{l}{\sqrt{z^2+l^2}})$$
OK, I see. yes, your starting equation is the correct one. ah, sorry about that. I forgot the displacement is from the equilibrium point of the spring, not the wall !

 

[BruceW]

OK, sorry about all that. I have one suggestion that is hopefully helpful. Since they don't explicitly say in the question to make $D<<l$ Then I think you should not make this approximation. Just keep the lower order terms in $\varepsilon$. And then at the end, you can always choose whether to make the approximation $D<<l$ or not. But since they don't explicitly say to use this approximation, my advice would be to not make this approximation.

 

[BvU]

@ehild: he did in post #1! and agreed to the minus sign for the k term in post #5.

All we worry about now is what to do with the $m\ddot{\varepsilon }+3k\frac{D^2ε}{l^2}=0$.

Chet is very quiet, so I'll chip in for him: for D we had (from $\epsilon = 0$, equilibrium): $k\frac{D^3}{l^2}=mg$. Well, his tip in post #7 is to be interpreted a little different from what skrat does in post #5. Chet wants to write $\ddot \epsilon + {3g\over D}\epsilon=0\$ so $\ \omega^2={3g\over D}$ from which the final answer $\omega\propto{1\over\sqrt{D}}\$ or $\ f = {1\over 2\pi} \sqrt{3g\over D}\$.

This is really nice, since for the much simpler case of one vertical spring with a weight the expression is the same, except the 3 is then a 1 !

At this point I wonder what part b of the exercise asks, since it wasn't worth mentioning because skrat already knew how to solve it...

 

[skrat]

I more or less got the same result in #5, but I agree that yours looks way better.

$\ddot{\varepsilon }+3k\frac{1}{ml^2}(\frac{mgl^2}{k})^{2/3}ε=0$
$\ddot{\varepsilon }+3k\frac{1}{ml^2}(D^3)^{2/3}ε=0$
$\ddot{\varepsilon }+3\frac{k}{ml^2}D^2ε=0$

We know that $D^3=\frac{gl^2m}{k}$ so $\frac{k}{ml^2}=\frac{g}{D^3}$

$\ddot{\varepsilon }+3\frac{g}{D^3}D^2ε=0$

$\ddot{\varepsilon }+\frac{3g}{D}ε=0$Part b is the same as here: https://www.physicsforums.com/showthread.php?p=4777170#post4777170

So part b wants me to write $z(t)$ if we release the mass from the same height where none of the two springs is deformed. It is the same problem as given on the link above.

 

[skrat]

HAHA now I see what bothers you. :D

The problem wants me to find $\omega (D)$ but I completely got rid of $D$ in #5 instead. Silly me. -.-

 

[BruceW]

@ehild: he did in post #1! and agreed to the minus sign for the k term in post #5.

I think ehild was replying to me, not to skrat.

p.s. I'm still concerned about using the approximation $D<<l$ Because even though they say to use this approximation in part b), if they don't say to use it in part a), then I think you shouldn't use it for part a).

 

[ehild]

The problem wants me to find $\omega (D)$ but I completely got rid of $D$ in #5 instead. Silly me. -.-

Yes, and find $\omega (D)$ by the Taylor expansion about D, as Bruce suggested.

ehild

 

[Chestermiller]

@ehild: he did in post #1! and agreed to the minus sign for the k term in post #5.

All we worry about now is what to do with the $m\ddot{\varepsilon }+3k\frac{D^2ε}{l^2}=0$.

Chet is very quiet, so I'll chip in for him: for D we had (from $\epsilon = 0$, equilibrium): $k\frac{D^3}{l^2}=mg$. Well, his tip in post #7 is to be interpreted a little different from what skrat does in post #5. Chet wants to write $\ddot \epsilon + {3g\over D}\epsilon=0\$ so $\ \omega^2={3g\over D}$ from which the final answer $\omega\propto{1\over\sqrt{D}}\$ or $\ f = {1\over 2\pi} \sqrt{3g\over D}\$.

This is not what I had in mind. What I had in mind was re-expressing Skrat's result in #5 as:
$$ω^2=3\left(\frac{kg^2}{ml^2}\right)^{1/3}$$
(i.e., without D present)

Chet

 

[BvU]

I am puzzled, Chet. How does this tell us how "the frequency of oscillation around equilibrium state depends on the initial offset due to the mass" ?
And: sorry for barging in with a wrong interpretation...

On the other hand, I liked the expression in post #13 but am of course always ready to yield for a better one.

@Bruce: they also don't mention the distance between the walls. The 3 shrivels back to 1/2 if that goes to zero. But so far only post #2 (ahem) mentioned that.

I think skrat is right in post #3: with 50 gram D is less than 1 cm. However, ehild carries a lot of weight (figuratively) in my book, so I am curious if she or someone else can come up with something sensible for bigger D/l . If I would take myself seriously (don't worry, I don't any more) that would change the 3 into something that varies from 3 for D=0 to 0.5 for D >> l , wouldn't it ?

 

[Chestermiller]

I am puzzled, Chet. How does this tell us how "the frequency of oscillation around equilibrium state depends on the initial offset due to the mass" ?
And: sorry for barging in with a wrong interpretation...

BvU, you're right. It doesn't. I guess I should have paid more attention to the problem statement. I fell in love with idea of expressing the result in terms of "l". But they were clearly looking for the result in terms of D. Without making the approximation that D<<l, I don't think it is possible to express the final result only in terms of D, without "l" also being present.

Chet

 

[ehild]

Read the original problem.

We have two springs each 0.5m long and both have k=5N/cm. One end of the spring is attached to the wall while the other two ends are connected in the middle. On the junction point of the two springs a mass is hanged. How does the frequency of oscillation around equilibrium state depend on the initial offset due to the mass?

Hint: There is no need to explicitly write the initial offset as function of mass. Just use the offset as independent parameter using which mass can be calculated.

In equilibrium, the springs are stretched because of the weight of the mass. D is the initial offset of the mass, below the origin, (O in the picture) . The mass is slightly displaced from the equilibrium position. Find the frequency of the small oscillations in terms of D.

Skrat found the differential equation for the motion of the mass.

$$\ddot z = g - \frac{2kz}{m}\left( 1-\left(1+\frac{z^2}{L^2}\right)^{-0.5}\right)$$

Expand the right side around z=D: Let z=D+ε. The right hand will have the form f(D)+(df/dz|_z=D)ε.

$$f(D)=g+\frac{2kD}{m}\left( \left(1+\frac{D^2}{L^2} \right)^{-0.5}-1 \right)$$

$$df/dz=\frac{2k}{m}\left( \left(1+\frac{D^2}{L^2} \right)^{-0.5}-1 -\left(1+\frac{D^2}{L^2} \right)^{-1.5}\frac{D^2}{L^2} \right)$$

Simplified: $$df/dz=\frac{2k}{m} \left( \left(1+\frac{D^2}{L^2} \right)^{-1.5} -1 \right )$$

f(D) has to be zero. If you know D, m can be determined.

$$ω^2=\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right )$$

I hope there is no error in the derivation. As I understood, use the approximation D/L<<1 when aswer to b).

ehild

 

[BruceW]

yeah, I get the same as ehild. And as ehild says, since f(D) is zero, this gives you an equation involving D,L and m. So if you are given some value for the mass, you could then in principle calculate D, and therefore calculate omega. Although the equation f(D)=0 is a little difficult to re-write for D, so this is probably why the question says you don't need to write the initial offset as a function of mass.

edit: Also interesting, is that if we now apply the approximation $D<<l$ then our first-order term goes to zero. So if you want to apply the approximation $D<<l$ Then you will need to keep at least second order terms in the Taylor expansion. This is not a big problem, but it just means you need to do a bit more differentiation. I don't know if this is important for part b) ? part b) seems like a different problem...

 

[ehild]

yeah, I get the same as ehild. And as ehild says, since f(D) is zero, this gives you an equation involving D,L and m. So if you are given some value for the mass, you could then in principle calculate D, and therefore calculate omega. Although the equation f(D)=0 is a little difficult to re-write for D, so this is probably why the question says you don't need to write the initial offset as a function of mass.

And the original problem suggest to express the mass in terms of D, which is much easier. And the result - ω in terms of D when D/L<<1 - is very simple an interesting.
Anyway, the frequency depends on the tension in the springs, and the tension is related to the offset, D.

 

[BruceW]

hehe, yeah, I've added a comment about the approximation in my edit. I guess he would need to keep higher orders in the Taylor expansion right? There would be no motion to first order, but there will be in the higher orders.

 

[BruceW]

This is a really nice problem in general. For the case of one spring, gravity just changes the equilibrium point, without changing the frequency. But in this case with two springs, gravity also changes the frequency. It's interesting, and kind of unexpected (for me at least). I like it when the answer to the problem is not just what I had expected it to be.

 

[BruceW]

edit: Also interesting, is that if we now apply the approximation $D<<l$ then our first-order term goes to zero. So if you want to apply the approximation $D<<l$ Then you will need to keep at least second order terms in the Taylor expansion. This is not a big problem, but it just means you need to do a bit more differentiation. I don't know if this is important for part b) ? part b) seems like a different problem...

ah, wait. what I said here is not right. It is more important to take $\varepsilon$ very small, than it is to take $D/l$ very small. So if you want to use the approximation $D<<l$ then you should not just let $D/l$ equal zero. But you can turn the equation:
$$ω^2=\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right )$$
into something much nicer, by letting $D/l$ be very small but not zero. So then, you still are just keeping the first-order term in $\varepsilon$.

 

[ehild]

Bruce, you explain things for yourself quite nicely. I meant that the equation first-order in ε is

$$\ddot \epsilon+\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right )\epsilon=0$$
with D and L and everyting else constant. It is an equation for SHM, with $$ω^2=\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right )$$

In case (D/L)²<<1,

$$ω^2≈\frac{2k}{m} \left( 1- (1-1.5\frac{D^2}{L^2}) \right )=\frac{3k}{m} \frac{D^2}{L^2}$$

Expanding f(D) $$f(D)=g+\frac{2kD}{m}\left( \left(1+\frac{D^2}{L^2} \right)^{-0.5}-1 \right)= g+\frac{2kD}{m} ( 1-0.5 \frac{D^2}{L^2} -1)\rightarrow g=\frac {kD^3}{mL^2}$$

But you can also isolate $\left(1+\frac{D^2}{L^2} \right)^{-0.5}$ from f(D)=0 and substitute it into the equation for ω² and expand after.

ehild

 

[BruceW]

Bruce, you explain things for yourself quite nicely. I meant that the equation first-order in ε is

$$\ddot \epsilon+\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right )\epsilon=0$$
with D and L and everyting else constant. It is an equation for SHM...

yeah... hehe I understand what you mean now. We must take the limit $\varepsilon \rightarrow 0$ first, before we think about taking a limit of D and L. (since we want the first-order motion, it is most important that $\varepsilon$ be small).

But you can also isolate $\left(1+\frac{D^2}{L^2} \right)^{-0.5}$ from f(D)=0 and substitute it into the equation for ω² and expand after.

ehild

ah, this is genius. you don't even need to expand after, it is already as simple as it could be, even without expanding it.

edit: ah no wait, I jumped the gun again. It is not as nice as it could be. The power is not the same. There is a power to 1.5 in the equation for frequency and a power to 0.5 in the f(D)=0 equation.

edit again: but, it does get rid of L, even without using the approximation. So that's pretty good. Although, to calculate D itself would require using L. But luckily, the question only asks for the frequency in terms of D.

 

[ehild]

yeah... hehe I understand what you mean now. We must take the limit $\varepsilon \rightarrow 0$ first, before we think about taking a limit of D and L. (since we want the first-order motion, it is most important that $\varepsilon$ be small).


ah, this is genius. you don't even need to expand after, it is already as simple as it could be, even without expanding it.

edit: ah no wait, I jumped the gun again. It is not as nice as it could be. The power is not the same. There is a power to 1.5 in the equation for frequency and a power to 0.5 in the f(D)=0 equation.

You can simplify... and then expand.


ehild

 

[BruceW]

yeah. we have
$$\left(1+\frac{D^2}{L^2} \right)^{-0.5} = 1 - \frac{gm}{2kD}$$
and the equation for frequency contains the cube of this. So then we just plug in the cube of the right hand side, into our equation for frequency, and take first order of the variable $gm/2kD$, since it will be small if $D/l$ is small.

Having said all this, I don't think the OP'er is meant to use the approximation $D/l<<1$. Even though it is an interesting approximation to make.

 

[BvU]

Wow, goes to show what an ambiguously formulated exercise can bring about. Why on Earth would someone ask something like that in such a way that in can be turned into such a hideous monstrum. You give l but don't give a clue as to whether it is anywhere near L.

I (sulk sulk) liked the $\ \ f = {1\over 2\pi} \sqrt{3g\over D}\$ a lot better. I now even have a hard time showing the b&e result goes to $\ \ f = {1\over 2\pi} \sqrt{g\over D}\$ for L going to zero...

 

[ehild]

In case (D/L)²<<1,

$$ω^2≈\frac{2k}{m} \left( 1- (1-1.5\frac{D^2}{L^2}) \right )=\frac{3k}{m} \frac{D^2}{L^2}$$

Expanding f(D) $$f(D)=g+\frac{2kD}{m}\left( \left(1+\frac{D^2}{L^2} \right)^{-0.5}-1 \right)= g+\frac{2kD}{m} ( 1-0.5 \frac{D^2}{L^2} -1)\rightarrow g=\frac {kD^3}{mL^2}$$

Divide the equations $$ω^2=\frac{3k}{m} \frac{D^2}{L^2}$$
and $$g=\frac {kD^3}{mL^2}$$

You get $$ω^2=\frac{3g}{D}$$.
...............
Using the original equation for ω²and f(D)=0, you can eliminate m before the approximation, and get a more accurate expression for ω². With the notation
$$η=\left(1+\frac{D^2}{L^2}\right)^{-0.5}$$:$$ω^2=\frac{2k}{m} \left( 1- \left(1+\frac{D^2}{L^2} \right)^{-1.5} \right ) \rightarrow$$
$$ω^2=\frac{2k}{m} ( 1- η^{3} )$$
and $$f(D)=g+\frac{2kD}{m}\left( \left(1+\frac{D^2}{L^2} \right)^{-0.5}-1 \right)=0\rightarrow g=\frac{2kD}{m}(1-η)$$
Dividing the equations, k and m cancel.

$$ω^2=\frac{g}{D}\frac{1-η^3}{1-η}=\frac{g}{D}(1+η+η^2)$$

With the approximation $η=1-0.5\frac{D^2}{L^2}$ when (D/L)²<<1,

$$ω^2=\frac{g}{D}(1+η+η^2)=\frac{3g}{D}(1-0.5\frac{D^2}{L^2})$$

With the given numerical data (L=0.5 m, k=500 N/m, m=0.05 kg), D≈0.063 m, (D/L)²≈ 0.016. Ignoring it means about half percent error in the frequency.

ehild