Questioning a Physics Problem: Taking Different Reference Frames

[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.

 

[simphys]

If someone sees this I wanted to ask an additonal question:
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?

 

[erobz]

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.

 

[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.

 

[Delta2]

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.

 

[kuruman]

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]

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...

 

[erobz]

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]

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.

 

[Orodruin]

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.

 

[kuruman]

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.

 

[erobz]

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

 

[erobz]

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]

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?

 

[erobz]

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 $U_{grav}$

 

[simphys]

my apologies.. I appreciate the help very much all!

 

[erobz]

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!

 

[Orodruin]

Thank you, I normally solve algebraically and sub in at the end.

That is usually the most advisable.

Speaking of dimensional analysis, you may find this to be of interest: https://www.physicsforums.com/insights/learn-the-basics-of-dimensional-analysis/

 

[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?)
thanks in advance

 

[simphys]

That is usually the most advisable.

Speaking of dimensional analysis, you may find this to be of interest: https://www.physicsforums.com/insights/learn-the-basics-of-dimensional-analysis/

thank you will check this out right now.

 

[erobz]

@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?)
thanks in advance

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.