Find v(x) of a mass suspended from a spring

[spsch]

Homework Statement

A mass (m=0.2 kg) is suspended from a spring (D/k=10 N/m)

When x = 0 (Spring is relaxed, at equilibrium), v(0) is 3 m/s.

Find v(x) as a first order differential equation.

Relevant Equations

I tried using energy conservation to check my work, but I think I messed up. I boxed in my solution. Above is me trying to verify with the conservation of energy and below is my attempt.

Hi all, I have a problem that I've been grappling with for the past 2 hours.

I was confident at first that I found the correct solution, but when I tried to verify I didn't have a constant in my v(x) function.

Here is my attempt:

I appreciate your help kind internet strangers!

 

[BvU]

I boxed in my solution

but someone else boxed 'We value your privacy' over your picture and I don't want to give them access to my computer.

 

[spsch]

What also throws me off is that they gave me k and m, but I didn't need it for my solution

 

[spsch]

but someone else boxed 'We value your privacy' over your picture and I don't want to give them access to my computer.

I'm sorry, I don't understand?

 

[BvU]

https://www.physicsforums.com/attachments/244545

 

[spsch]

https://www.physicsforums.com/attachments/244545

I'm sorry I can't open your attachement, page not found.

 

[spsch]

https://www.physicsforums.com/attachments/244545

Do you get a cookie notification from imgur? Meaning I should repost my picture?
I try uploading it as attachment.

 

[BvU]

I'm sorry I can't open your attachement, page not found.

Don't know what went wrong:

 

[BvU]

What also throws me off is that they gave me k and m, but I didn't need it for my solution

You need $k$ and $m$ in the differential equation you are supposed to set up -- at least according to your own problem statement:

Find v(x) as a first order differential equation.

 

[haruspex]

didn't have a constant in my v(x) function.

Please clarify: which equation and which constant do you think you are missing?
If you mean your differential equation (the one with vdv/dx) and the constant C, you would not expect that constant in that equation. The given information about the value of v at x=0 is redundant.

 

[spsch]

You need $k$ and $m$ in the differential equation you are supposed to set up -- at least according to your own problem statement:

Thank you. Were you able to look at my attempt? Does the differential look right?
https://www.physicsforums.com/attachments/img_20190603_172630-jpg.244546/

 

[spsch]

Please clarify: which equation and which constant do you think you are missing?
If you mean your differential equation (the one with vdv/dx) and the constant C, you would not expect that constant in that equation. The given information about the value of v at x=0 is redundant.

Hi, can you see my attachement https://www.physicsforums.com/attachments/img_20190603_172630-jpg.244546/ (i am asking because I seem not to be able to open all)

When I solve for v(x) with differential equations and when I use conservation of Energy I get two solutions that differ by a constant 9 under the square root.

 

[spsch]

The function itself is also weird shouldn't v(x) eventually come to 0 when kx = mg?

 

[haruspex]

When I solve for v(x) with differential equations and when I use conservation of Energy I get two solutions that differ by a constant 9 under the square root.

I was misled by the layout of your working. I thought you were comparing the boxed equation half way down with the final one.
Right at the start I see you have $\frac 12kx^2=mgh+\frac 12mv^2$.
Please explain the basis of this equation and what h represents there.
Soon after, you quietly dropped the minus sign in $\sqrt{-2gh}$.

shouldn't v(x) eventually come to 0 when kx = mg?

Why? If that condition arises it would then be static.

 

[spsch]

I was misled by the layout of your working. I thought you were comparing the boxed equation half way down with the final one.
Right at the start I see you have $\frac 12kx^2=mgh+\frac 12mv^2$.
Please explain the basis of this equation and what h represents there.
Soon after, you quietly dropped the minus sign in $\sqrt{-2gh}$.

Why? If that condition arises it would then be static.

Hi Haruspex.
Yes, I'm sorry the layout is bad I see that now. I had the work on another sheet and I actually thought it would be more clear this way.

The above is me trying to verify my solution by using conservation of energy.

This is what I thought:
At the middle where v is at it's maximum and x = 0 there is only potential energy mgh (or mgy) and the kinetic energy which is at it's maximum there.
On the way down the spring force works against the kinetic energy and gravity, i.e. pulls the mass up.

if we say that h= 0 where kx = mg then all the energy there, at the bottom so to say is(1/2)*k*x^2
$\frac 12kx^2$. So all of the potential energy and kinetic energy turned into potential spring energy.I dropped the minus sign because g is negative (-)(-) = (+), but I've wondered about whether I made a mistake here myself already, I actually dreamt about it.I thought at the bottom v should be zero for a short time and reverse. I was imagining it.

 

[BvU]

I thought at the bottom v should be zero for a short time and reverse. I was imagining it

I don't think you imagine it:

(picture from Wikipedia)

Please confirm that you are searching for a first-order differential equation for $v$, as your post #1 indicated...

(because usually we deal with this harmonic oscillator as a second-order differential equation in $x$)

 

[spsch]

Hi BvU

Thank you for replying.

This is exactly what I was imagining it to look like (your gif animation).

I am looking for first-order differential equation.

The exact problem is two parts, a and b.

a) Find a 1st Order differential Equation Expression for v(x).

b) Solve the Equation you found for v(x)

it's from an old swiss high school physics exam.

 

[haruspex]

At the middle where v is at it's maximum and x = 0

There is an apparent conflict in the given information. It says "When x = 0 (Spring is relaxed, at equilibrium)", but the equilibrium position (with the mass attached) is not with the spring relaxed.
Anyway, I'll assume it means equilibrium.
But your energy equation is still wrong. The total energy is constant. Rewrite the equation to express that.

 

[spsch]

There is an apparent conflict in the given information. It says "When x = 0 (Spring is relaxed, at equilibrium)", but the equilibrium position (with the mass attached) is not with the spring relaxed.
Anyway, I'll assume it means equilibrium.
But your energy equation is still wrong. The total energy is constant. Rewrite the equation to express that.

Hi Haruspex,
I am sorry!
It actually only says at x=0 where the spring is relaxed v = 3 m/s. Assumed that this must be equilibrium and attempted to translate.

There is a diagram with it. I attached a screenshot.

I will redo the equation and see if I can get the two to mean the same thing!

thank you very much.

there also is a hint with a = v*dv/dx, which is why I went with kx-mg=mv*dv/dx

 

[haruspex]

It actually only says at x=0 where the spring is relaxed v = 3 m/s.

Ok, but try to write the energy conservation equation now.

 

[spsch]

Ok, but try to write the energy conservation equation now.

Thank you, I am, I'm trying to figure out where I'm implying that it isn't constant. I had thought of the way that I wrote it that it expresses that the total energy is constant.
:-S.

 

[haruspex]

Thank you, I am, I'm trying to figure out where I'm implying that it isn't constant. I had thought of the way that I wrote it that it expresses that the total energy is constant.
:-S.

You have spring PE on one side and KE on the other, both positive. E.g. both can increase with no change to the GPE.

 

[spsch]

You have spring PE on one side and KE on the other, both positive. E.g. both can increase with no change to the GPE.

Hello Haruspex,

Thank you very much for your help.
I gave it a lot of thought, and by myself, I still didn't quite get what's wrong.

But by your hint, I think I may have figured out what the right equation is.
I rewrote it and uploaded it as an attachment, together with a drawing which attempts to explain my thinking in how I got the first equation.
Would you mind checking if the new equation is correct?

I changed h to be A-x, where A is the amplitude.

I am sorry if I'm a bit slow, I haven't had formal schooling, and I found that I sometimes taught myself things the wrong way when it's way too late. But that's why I also really appreciate the help, like having a teacher!

 

[haruspex]

Hello Haruspex,

Thank you very much for your help.
I gave it a lot of thought, and by myself, I still didn't quite get what's wrong.

But by your hint, I think I may have figured out what the right equation is.
I rewrote it and uploaded it as an attachment, together with a drawing which attempts to explain my thinking in how I got the first equation.
Would you mind checking if the new equation is correct?

I changed h to be A-x, where A is the amplitude.

I am sorry if I'm a bit slow, I haven't had formal schooling, and I found that I sometimes taught myself things the wrong way when it's way too late. But that's why I also really appreciate the help, like having a teacher!

It's rather hard to follow your working without some explanation. Don't worry about the amplitude for now. That is a "boundary condition" we can plug in later if appropriate.
You are defining x as displacement from spring’s relaxed position. I'll assume up is positive.
At x, you have that the elastic PE (EPE) is $\frac 12kx^2$ and that the KE is $\frac 12mv^2$
What is the GPE (relative to x=0)? So what expression do you get for the total energy at x?

Next, plug in the known values at x=0 to find the value of that constant energy.

 

[SammyS]

Hi all, I have a problem that I've been grappling with for the past 2 hours.

I was confident at first that I found the correct solution, but when I tried to verify I didn't have a constant in my v(x) function.

In your Original Post you had the following .

So that first line refers to the total energy at $x=0$, at which location the spring is unstretched. The gravitational potential here is referenced to some other point by $h$.

The second line appears to be the total energy at some value of $x$ at which the gravitational potential is zero and the velocity, $v$, is also zero. Since the gravitational potential here is zero, this must be a distance, $h$ below $x=0$ and $h$ must be positive, i.e. it must be that at this location $x=-h$.

Thus the gravitational potential at an arbitrary location is given by $mg(x+h)$.

 

[spsch]

Thank you haruspex, I think I see my mistake thanks to you and sammy, i'll try to rewrite it now.

 

[spsch]

It's rather hard to follow your working without some explanation. Don't worry about the amplitude for now. That is a "boundary condition" we can plug in later if appropriate.
You are defining x as displacement from spring’s relaxed position. I'll assume up is positive.
At x, you have that the elastic PE (EPE) is $\frac 12kx^2$ and that the KE is $\frac 12mv^2$
What is the GPE (relative to x=0)? So what expression do you get for the total energy at x?

Next, plug in the known values at x=0 to find the value of that constant energy.

It took me an awful many attempts, but I think I almost got it now. I am so grateful. Posting in a few!

 

[spsch]

In your Original Post you had the following .
View attachment 244602
So that first line refers to the total energy at $x=0$, at which location the spring is unstretched. The gravitational potential here is referenced to some other point by $h$.

The second line appears to be the total energy at some value of $x$ at which the gravitational potential is zero and the velocity, $v$, is also zero. Since the gravitational potential here is zero, this must be a distance, $h$ below $x=0$ and $h$ must be positive, i.e. it must be that at this location $x=-h$.

Thus the gravitational potential at an arbitrary location is given by $mg(x+h)$.

Thank you! I think I have found a solution now! And it seems to verify my differential.

 

[spsch]

I think this is it? I graphed it and it looks like what I imagined the graph to be like that there are two zeros (when spring is all the way up and when it is all the way at -h).

There is a discrepancy with the negative sign to my differential though.

should kx-mg = ma actually be -kx-mg = ma?

Does this otherwise look correct?

Thank you so much, really!

 

[haruspex]

I think this is it? I graphed it and it looks like what I imagined the graph to be like that there are two zeros (when spring is all the way up and when it is all the way at -h).

There is a discrepancy with the negative sign to my differential though.

should kx-mg = ma actually be -kx-mg = ma?

Does this otherwise look correct?

Thank you so much, really!

Don't worry about h and v₀ yet. Start with the general condition as in post #24. If it is at x above the relaxed position and moving upwards at speed v:
What is the spring PE?
What is the KE?
What is the GPE (relative to x=0)?
What is the total?

 

[spsch]

Thank you. I did follow these questions which had led me to that result.

I think:

The spring PE is 0 at x=0

The KE is (1/2)*m*v^2

If we take x=0 as the height being 0 then GPE is 0

So the total then would be just the KE? (1/2)*m*v^2

 

[SammyS]

Thank you. I did follow these questions which had led me to that result.

I think:

The spring PE is 0 at x=0

The KE is (1/2)*m*v^2

If we take x=0 as the height being 0 then GPE is 0

So the total then would be just the KE? (1/2)*m*v^2

Yes, then the GPE is 0 at x = 0.

What is the GPE at an arbitrary value for x ?

 

[haruspex]

The spring PE is 0 at x=0

I didn't ask about x=0. My questions are in the context of an arbitrary value of x.

 

[spsch]

Thank you. I did follow these questions which had led me to that result.

Yes, then the GPE is 0 at x = 0.

What is the GPE at an arbitrary value for x ?

x+h? as h being the displacement from x=0?

 

[spsch]

I didn't ask about x=0. My questions are in the context of an arbitrary value of x.

Oh, then 1/2*k*x^2?

I'm sorry I may have misunderstood.

 

[SammyS]

Thank you. I did follow these questions which had led me to that result.

x+h? as h being the displacement from x=0?

For one thing, x itself is the displacement from the location, x=0.
(If you are taking GPE to be zero at x=0, then h should not enter into this anymore.)

I asked for the GPE (gravitational potential energy) of the mass when it's at location x .

This should help you answer @haruspex's question(s)

 

[spsch]

Oh! of course, so just mgx?

 

[haruspex]

Oh! of course, so just mgx?

Right, so what is the total mechanical energy at height x?

 

[spsch]

Right, so what is the total mechanical energy at height x?

(1/2)*m*v^2 + mgx+(1/2)*k*x^2?

 

[haruspex]

(1/2)*m*v^2 + mgx+(1/2)*k*x^2?

Good. And this should be constant, and we know all the values at t=0. So what general equation does that give you?

 

[spsch]

(1/2*m*v₀^2) = (1/2)*m*v^2 + mgx+(1/2)*k*x^2

$v = \sqrt{ v_^2 - \frac {kx^2} {m}, - 2gx}$ (/ I'm sorry tried to but won't convert)

v = sqrt( v₀^2 - (k*x^2) /m), - 2*g*x)

could this be it?

 

[haruspex]

(1/2*m*v₀^2) = (1/2)*m*v^2 + mgx+(1/2)*k*x^2

$v = \sqrt{ v_0^2 - \frac {kx^2} {m} - 2gx}$

could this be it?

That's it. (Latex doesn't like [ SUB] etc. use _)

Can you turn that into t= an integral wrt x?

 

[spsch]

That's it. (Latex doesn't like [ SUB] etc. use _)

Hmm, I subbed the sub with _ but it shows a square now, but i'll figure this out too :-)

Can you turn that into t= an integral wrt x?

Hmm, with t? I think so, I'm anxious to see what solution I get. I'll try and get back.

PS, I am grateful that you made me work through the problem, even though it's a small difference, I now really understand what I did wrong. I don't think I would have this insight if you'd just told me to switch x with h. Thank you very much!

 

[BvU]

I'm sorry tried to but won't convert

also: put a \ before the sqrt: $$v = \sqrt{ v_0^2 - \frac {kx^2} {m} - 2gx} $$


(you could 'check' this expression: when is $v=0$ and when is it at a maximum ?)

And this is not the whole 'it', but a step towards the differential equation that the exercise asks ...

My impression is that it wants an equation in the form ${dv\over dx} =$ { an expression in $x$ }
$\$

 

[haruspex]

also: put a \ before the sqrt: $$v = \sqrt{ v_0^2 - \frac {kx^2} {m} - 2gx} $$


(you could 'check' this expression: when is $v=0$ and when is it at a maximum ?)

And this is not the whole 'it', but a step towards the differential equation that the exercise asks ...

My impression is that it wants an equation in the form ${dv\over dx} =$ { an expression in $x$ }
$\$

Ah yes, @spsch, I was forgetting the original purpose of the thread. You were just checking an answer by using energy. So to get a differential equation involving v from your post #41 you now need to differentiate. But it is not clear to me whether you need to differentiate wrt t or wrt x. It depends what ODE you got that you are trying to verify.

 

[spsch]

also: put a \ before the sqrt: $$v = \sqrt{ v_0^2 - \frac {kx^2} {m} - 2gx} $$


(you could 'check' this expression: when is $v=0$ and when is it at a maximum ?)

And this is not the whole 'it', but a step towards the differential equation that the exercise asks ...

My impression is that it wants an equation in the form ${dv\over dx} =$ { an expression in $x$ }
$\$

Thank you! the \sqrt wasn't enough yet. Don't worry about this though, you guys do enough already!

Probably a dumb question but... wouldn't dv/dx just be the derivative of the right side?

 

[spsch]

Ah yes, @spsch, I was forgetting the original purpose of the thread. You were just checking an answer by using energy. So to get a differential equation involving v from your post #41 you now need to differentiate. But it is not clear to me whether you need to differentiate wrt t or wrt x. It depends what ODE you got that you are trying to verify.

Hi Haruspex. I'm not sure what you mean with what ODE?

I had a solution which is actually what I got now with your help.

I got that by setting kx-mg = ma and using v*dv/dx for a as it was a hint on the problem.

The problem was looking for v(x)

 

[spsch]

So I got this as the derivative. Is this a(x)?

 

[BvU]

\sqrt wasn't enough yet

$\TeX$ now stumbles over _^ without anything in between

v_0^2 will fix it. You also don't really want the comma. And don't worry: we do it because we like to do it or else we wouldn't do it

Given up on $\LaTeX$ ? and back to images...

Methinks the image in #48 is a good answer to the problem statement in #19 (also an image ).
Whether it's still correct can be checked 'easily':

• ${dv\over dx} = 0\ \$ at $\ \ x_0 = - {mg\over k}$
• substitute $\omega =\sqrt{k\over m}, \ \ \xi = x_0-x, \ \ v =$ ... oh, well, maybe not that easy ...

and all that for 3 points ...

Makes me curious to see parts b) and c)
$\$

 

[BvU]

So I got this as the derivative. Is this a(x)?

I should think $a(x) = -\omega^2 (x-x_0) \ \$ with $\ \ \omega^2 = {k\over m}$

In other words: ${da\over dx} = -\omega^2$