# Speed of a mass is not correct when taking certain reference frames for PE's

simphys
Homework Statement:
II) A vertical spring (ignore its mass), whose spring
constant is is attached to a table and is
compressed down by 0.160 m.
(a) What upward speed can
it give to a 0.380-kg ball when released? (b) How high
above its original position (spring compressed) will the
ball fly?
Relevant Equations:
E1 = E2
for (a): I basically got the correct answer, but when resolved with taking different reference lines/frames I got a different answer.

for the 1st attempt I took y = 0 (for both ##U_{el}## and ##U_{grav}) at the position where the spring is uncompressed.
for the 2nd attempt (with wrong solution) I took y = 0 for ##U_{grav}## and x = 0 for ##U_{el}## as is shown on the following picture.
have I made a mistake or is this simply not possible? if it is not clear, please tell me.

• Delta2

simphys
A week ago I started studying chemistry (as a beginner) where the 1st chapter covered Dimensional analysis and the scientific method. Before this treatment of DA i didn't really deem it necessary to use DA while solving a problem and to be honest when doing problems (as wel as for mechanics of materials etc.) I really got a loooot of 'small' stupid errors.
So my question is.. Can I consider dimensional analysis as the 'way' to check whether your solution is 'correct' in terms of formula's used and thus minimizing the errors?

Gold Member
Is that a picture of attempt (1) or (2). I love a good diagram, and this is not a good diagram nor is it a particularly good explanation of what is shown. Present the first solution too that you think is correct.

It's also better if you save the pictures for diagrams and show your math in Latex. It makes a cleaner read for a potential helper. Also, doing these things (while may seem time consuming) helps you to slow down and organize your thoughts, which can often help you to find the discrepancy on your own.

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• Delta2 and simphys
simphys
for the 2nd attempt (with wrong solution) I took y = 0 for Ugrav and x = 0 for Uel as is shown on the following picture.
for the second one
Is that a picture of attempt (1) or (2). I love a good diagram, and this is not a good diagram nor is it a particularly good explanation of what is shown. Present the first solution too that you think is correct.

It's also better if you save the pictures for diagrams and show your math in Latex. It makes a cleaner read for a potential helper. Also, doing these things (while may seem time consuming) helps you to slow down and organize your thoughts, which can often help you to find the discrepancy on your own.
You didn't like my diagram? :(
How could I improve upon it?
and what do you mean not a good explanation?
And Yeah you're right on that no need to rush, best to do a couple of ex than the whole selection lol.

I will do that. To be honest, I normally put ? with unknowns and ! with knowns and then solve. not on this one for some reason.

Homework Helper
Gold Member
I also can't understand where you took your zeroth level of potential energy for the gravitational and spring potential energy. if I understand correctly however, when you took different zeroth levels for each of the PEs then you didn't get the expected result ,right?

The natural choice ( or at least what I would do to solve this problem) is to get the zeroth level of PE for both where the spring is uncompressed.

• erobz
Homework Helper
Gold Member
So my question is.. Can I consider dimensional analysis as the 'way' to check whether your solution is 'correct' in terms of formula's used and thus minimizing the errors?
Absolutely. Dimensional analysis is the first check when you derive an answer in symbolic form. If you end up with an answer that is dimensionally incorrect, you know immediately that it can't be correct. Then you can backtrack end check the equations that led to the bottom line individually to pinpoint the source of error, pehaps forgetting to take a square root, or whatever. Of course DA doesn't help with factors of 2 or π, but it is a good starting point for troubleshooting your work.

Now regarding the original post. Evidently you want to see what happens when you choose a different zero for the potential energy. Frankly, I too do not fully understand what you did but this is what I would do.

First I define an arbitrary origin of coordinates on a one dimensional axis
Let ##x_0=## the position of the tip of the spring that is free to move when it is relaxed.
Let ##x=## the general coordinate of the tip of the spring that is free to move.

The force that the spring exerts as a function of the tip's position is ##F(x)=-k(x-x_0)##
The potential energy function is, by definition, \begin{align} & U(x)=-\int_{x_{\text{ref}}}^xF(x)dx=+k \int_{x_{\text{ref}}}^x(x-x_0)dx=\frac{k x^2}{2}-k x {x_0}+k {x_0} x_{\text{ref}}-\frac{kx_{\text{ref}}^2}{2}\nonumber \\& U(x)=\frac{1}{2}k\left(x^2-x_{\text{ref}}^2\right)-kx_0(x-x_{\text{ref}}).\nonumber \end{align}This expression allows different choices for the origin and the reference of the potential energy. It is convenient to set the origin of coordinates and the reference of the potential at the tip of the spring, i.e. ##x_0=0##, and ##x_{\text{ref}}=0## in which case the potential energy function of the spring is the familiar $$U(x)=\frac{1}{2}kx^2.$$

• simphys
simphys
I'm on the phone right now, but I will put my solution in latex when I ask the next question, I will rewrite the solution for the probelm (in both ways in a second.)

simphys
Absolutely. Dimensional analysis is the first check when you derive an answer in symbolic form. If you end up with an answer that is dimensionally incorrect, you know immediately that it can't be correct. Then you can backtrack end check the equations that led to the bottom line individually to pinpoint the source of error, pehaps forgetting to take a square root, or whatever. Of course DA doesn't help with factors of 2 or π, but it is a good starting point for troubleshooting your work.

Now regarding the original post. Evidently you want to see what happens when you choose a different zero for the potential energy. Frankly, I too do not fully understand what you did but this is what I would do.

First I define an arbitrary origin of coordinates on a one dimensional axis
Let ##x_0=## the position of the tip of the spring that is free to move when it is relaxed.
Let ##x=## the general coordinate of the tip of the spring that is free to move.

The force that the spring exerts as a function of the tip's position is ##F(x)=-k(x-x_0)##
The potential energy function is, by definition, \begin{align} & U(x)=-\int_{x_{\text{ref}}}^xF(x)dx=+k \int_{x_{\text{ref}}}^x(x-x_0)dx=\frac{k x^2}{2}-k x {x_0}+k {x_0} x_{\text{ref}}-\frac{kx_{\text{ref}}^2}{2}\nonumber \\& U(x)=\frac{1}{2}k\left(x^2-x_{\text{ref}}^2\right)-kx_0(x-x_{\text{ref}}).\nonumber \end{align}This expression allows different choices for the origin and the reference of the potential energy. It is convenient to set the origin of coordinates and the reference of the potential at the tip of the spring, i.e. ##x_0=0##, and ##x_{\text{ref}}=0## in which case the potential energy function of the spring is the familiar $$U(x)=\frac{1}{2}kx^2.$$
Thank you for the explanation! I have one questoin however, what would be the difference between ##x_{ref}## and ##x_{origin}##

and I had however a little bit of a different question, I will show the solution in a second once I rewrite the solutions.

And for the 1st part:
Thanks a lot. It really feels as that 'uncertainty' that I always had in my answer is kind of gone. just by checking the dimensions tbh.
Before that I was always rushing to make the most exercises and DA and check up steps + wrote way too messy to the point where you can't easily retrace your mistake and just again without knowing what mistake you made lo.l...

Gold Member
I'm on the phone right now, but I will put my solution in latex when I ask the next question, I will rewrite the solution for the probelm (in both ways in a second.)
In the diagrams just try to be as clear as possible in labeling the various positions and datums. Ideally diagram for solution 1 - supporting math, diagram for solution 2 - supporting math. Also, write down steps that you think are taken for granted. Basically, go out of your way to help people to help you and you'll have a better overall experience, I think.

• simphys
simphys
In the diagrams just try to be as clear as possible in labeling the various positions and datums. Ideally diagram for solution 1 - supporting math, diagram for solution 2 - supporting math. Also, write down steps that you think are taken for granted. Basically, go out of your way to help people to help you and you'll have a better overall experience, I think.
Oh great idea, thank you!
And yep I agree, and not only that, it's also easier for yourself to check what you have (possibly done wrong).
But thank you for the idea it'll be even more organized that way.

• erobz
Staff Emeritus
Homework Helper
Gold Member
Dimensional analysis is the first check when you derive an answer in symbolic form.
… and carrying units through the computation is a good dimensional check when doing things numerically.

Homework Helper
Gold Member
Thank you for the explanation! I have one questoin however, what would be the difference between ##x_{ref}## and ##x_{origin}##
They can have different values. For example, you can pick your origin at the fixed end of a spring of relaxed length ##L##, in which case ##x_0=L## and the zero of potential energy when the spring is compressed to half its length ##x_{\text{ref}}=L/2## or whatever other values you please. The most convenient choice is at the free tip of the spring for both because the resulting expression is easy to remember.

simphys
I also can't understand where you took your zeroth level of potential energy for the gravitational and spring potential energy. if I understand correctly however, when you took different zeroth levels for each of the PEs then you didn't get the expected result ,right?

The natural choice ( or at least what I would do to solve this problem) is to get the zeroth level of PE for both where the spring is uncompressed.
sorry missed your comment, that is exactly it!
And before that I used the same datum line for ##U_{grav}## and ##U_{el}## (just the y-coordinate no x as a matter a fact) and got 7.7 m/s

simphys
They can have different values. For example, you can pick your origin at the fixed end of a spring of relaxed length ##L##, in which case ##x_0=L## and the zero of potential energy when the spring is compressed to half its length ##x_{\text{ref}}=L/2## or whatever other values you please. The most convenient choice is at the free tip of the spring for both because the resulting expression is easy to remember.
oh yeah right.. okay that clears that one totally up! Thank you.

simphys
This is the solution that is not correct by using two different points of reference for U_grav and U_el
And do they need to be the same? simphys
… and carrying units through the computation is a good dimensional check when doing things numerically.
Thank you, I normally solve algebraically and sub in at the end.

Gold Member
You took ## x_1## to be ##0## from what appears to be measured from ##x_2##, but it's in the computation. In effect you seem to be saying the potential energy of the spring in the diagram labed ##(1)## is ##0##. Is that correct?

simphys
You took ## x_1## to be ##0##, but its in the computation. In effect you seem to be saying the potential energy of the spring in the diagram labed ##(1)## is ##0##. Is that correct?
no ##x_1## is 0.16m because ##1## is the initial possition

simphys
Well look at like this,
I used the 'most' convenient points for both to be 0 at that point
i.e.
for U_EL x = 0 at uncompressed spring
for U_grav y = 0 at lowest point aka compressed position

Gold Member
Well look at like this,
I used the 'most' convenient points for both to be 0 at that point
i.e.
for U_EL x = 0 at uncompressed spring
for U_grav y = 0 at lowest point aka compressed position
Ok, you are saying that:

$$\frac{1}{2}k x_1^2 = mgy_2 + \frac{1}{2} m v_2^2$$

is incorrect?

• simphys
simphys
Ok, you are saying that:

$$\frac{1}{2}k x_1^2 = mgy_2 + \frac{1}{2} m v_2^2$$

is incorrect?
presumably, yes or those datum lines are just not permitted for some reason Is my assumption?

Gold Member
presumably, yes or those datum lines are just not permitted for some reason Is my assumption?
Hmmm. I disagree.

simphys
Hmmm. I disagree.
Oh my god I am going to cry...

simphys
Hmmm. I disagree.
I am sorry... my initial equation was simply wrong aka the solution of 7.7 m/s because I forgot to include the ##U_{grav}##

• Delta2
simphys
my apologies.. I appreciate the help very much all!

• erobz and Delta2
Gold Member
I am sorry... my initial equation was simply wrong aka the solution of 7.7 m/s because I forgot to include the ##U_{grav}##
Hence the importance of providing the "correct" solution for people to examine as well!

• simphys
Staff Emeritus
Homework Helper
Gold Member
• simphys
simphys
@erobz
I had one more question for you about looking at it as situation per situation.
In the problem as stated velow.
(II) A pendulu(II) A pendulum 2.00 m long is released (from rest) at an angle
##/theta_0##=30degrees (Fig. 14). Determine the speed of the 70.0-g bob: (a) at the lowest point (b) at ##/theta = 15##(c) at ##/theta = -15## (i.e., on the opposite side). (d) Determine the tension in the cord at each of these three points. (e) If the bob is given an initial speed ##v_0## when released at #theta_0 = 30## degrees recalculate the speeds for parts (a), (b), and (c)

Would you
solve all the parts seperately or would you f.e. solve everything related to the situation first and then move on to the next situation (that is first ##/theta = 0## and then ##/theta = 15## f.e. or (a) then (b) and so on?)

Gold Member
@erobz
I had one more question for you about looking at it as situation per situation.
In the problem as stated velow.
(II) A pendulu(II) A pendulum 2.00 m long is released (from rest) at an angle
##/theta_0##=30degrees (Fig. 14). Determine the speed of the 70.0-g bob: (a) at the lowest point (b) at ##/theta = 15##(c) at ##/theta = -15## (i.e., on the opposite side). (d) Determine the tension in the cord at each of these three points. (e) If the bob is given an initial speed ##v_0## when released at #theta_0 = 30## degrees recalculate the speeds for parts (a), (b), and (c)

Would you
solve all the parts seperately or would you f.e. solve everything related to the situation first and then move on to the next situation (that is first ##/theta = 0## and then ##/theta = 15## f.e. or (a) then (b) and so on?)
I think per Homework Forum Rules you have to try it before we help. I think the reason for this is that outlining an approach for you, is depriving you of developing that important part of the problems solution.

• Delta2 and simphys
simphys
I think per Homework Forum Rules you have to try it before we help. I think the reason for this is that outlining an approach for you, is depriving you of developing that important part of the problems solution.
What would you consider the important part of the problems solution? Figuring it out on your own I or?
But no not at all, I will show you the solution, I've got it here :) I totally understand it and stuff.
But What I want to do is to improve the way of how I structure the problems. Because I tended to rush things through and also sometimes mid-problem got stuck simply because of not being able to recognize my past writing from what is for what aka (problems statics and strength of materials combined for example)

You see because I've never had science (or ... barely any) only biology actually 2h/week I am not very accustomed to following a structure that's clear and easily retracable for mistakes f.e.. If I showed to problems from the past you would NOT comprehend a single part if I am being honest.
Of course the books outlines an 'approach' but it's different you know.
And thus.. that is what I want to improve upon

Gold Member
I don't know, while it's tied to conservation of energy it is a completely different problem. It probably deserves its own thread. I'd suggest starting a new one.

(1) Practicing making/presenting good diagrams and supporting math in LaTeX is good practice, and practice makes perfect. If you need practice in exercising patience and effectively communicating scientific ideas, then that is the way to do it.

(2) It's not until you try to organize the problem and explain it to others (maybe many times over) that more efficient approaches become clear.

(1) and (2) are both important parts of the problem-solving process.

If you want to improve you must practice.

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• Delta2 and simphys
simphys
I don't know, while it's tied to conservation of energy it is a completely different problem. It probably deserves its own thread. I'd suggest starting a new one.

(1) Practicing making/presenting good diagrams and supporting math in LaTeX is good practice, and practice makes perfect. If you need practice in exercising patience and effectively communicating scientific ideas, then that is the way to do it.

(2) It's not until you try to organize the problem and explain it to others (maybe many times over) that more efficient approaches become clear.

(1) and (2) are both important parts of the problem-solving process.

If you want to improve you must practice.
Thanks a lot for the advice! Will definitely do that from now, and for now I can only say those diagrams oh man.. to describe energy situations it really helps you visualize it instead of just thinking about it in your head.

anyways thank you

• erobz
Gold Member
Thanks a lot for the advice! Will definitely do that from now, and for now I can only say those diagrams oh man.. to describe energy situations it really helps you visualize it instead of just thinking about it in your head.

anyways thank you
"Through hardship comes enlightenment"

• Delta2 and simphys
simphys
"Through hardship comes enlightenment"
oh 100% we love that :P

• erobz and Delta2