Dimensional analysis problem

In summary: If we assume that the amplitude is also a function of b (and not just the displacement), we get a nonsensical result.The second assumption is that the force is a restoring force.This is problematic because it is only valid for a negative displacement. If the displacement is greater than zero, the force is no longer a restoring force.
  • #1
117
1

Homework Statement



hi I am trying out a question on dimensional analysis , I am having trouble finding the solution when the force is non linear , the situation is a mass resting on a smooth surface attatched to a spring , the spring is attached to a wall on its other end ,


Homework Equations


the expression for the non linear force is
[tex] f= -kx + bx^2[/tex]


The Attempt at a Solution


, what will be the expression for the frequency? [tex]\omega=m^d *k^e* (bx)^f*x^g[/tex]? or maybe not since its not giving me the expected answer. i found the expression for the case when the force is linear [itex]f=-kx[/itex]: [itex]\omega= c \sqrt{k/m}[/itex] , the epected answer for the non linear case is of the form [tex] \omega = f(x_o \frac{b}{k}) \sqrt{k/m} [/tex]
 
Physics news on Phys.org
  • #2
If I understand this correctly, then you're looking for the frequency of the motion up to multiplication by a constant, right?

Well, your approach here should be purely one of dimensional analysis, without referring to the analog of SHM.

The one thing you can say about the frequency is that it is independent of the displacement, as such, you expect it to be a function of the parameters of your system.

[tex]\omega = f(m, k, b)[/tex] with dimensions of [tex][time]^{-1}[/tex] (This approach doesn't give a result since you're stuck with the length dimension of [tex]b[/tex])

Another way to approach it would be to assume that it is some multiple of the frequency of the linear system, [tex]\omega _0 = \sqrt{\frac{k}{m}}[/tex] which could prove fruitful here. (The same problem as above persists)

I think the insight required here is to notice that the force is a restoring force only for [tex]-x_c < x <x_c[/tex] where [tex]x_c=\frac{k}{b}[/tex]

And then using that as a clue to incorporate [tex]x_c[/tex] in your solution.

Is the expected answer you posted the textbook's answer? I'm asking since it makes absolutely no sense.. If we are to understand that [tex]x_0=x_c[/tex], then [tex]f(x_0 \frac{b}{k})=f(1)[/tex] and that doesn't make sense dimensionally. But let's assume that the meaning was just 1, then all you get is just the linear frequency.

Where did you get this question..?
 
Last edited:
  • #3
RoyalCat said:
If I understand this correctly, then you're looking for the frequency of the motion up to multiplication by a constant, right?
yes your right,
RoyalCat said:
Well, your approach here should be purely one of dimensional analysis, without referring to the analog of SHM.
i hope i did not refer to SHM,

RoyalCat said:
The one thing you can say about the frequency is that it is independent of the displacement, as such, you expect it to be a function of the parameters of your system.

[tex]\omega = f(m, k, b)[/tex] with dimensions of [tex][time]^{-1}[/tex] (This approach doesn't give a result since you're stuck with the length dimension of [tex]b[/tex])

Another way to approach it would be to assume that it is some multiple of the frequency of the linear system, [tex]\omega _0 = \sqrt{\frac{k}{m}}[/tex] which could prove fruitful here. (The same problem as above persists)

I think the insight required here is to notice that the force is a restoring force only for [tex]-x_c < x <x_c[/tex] where [tex]x_c=\frac{k}{b}[/tex]

And then using that as a clue to incorporate [tex]x_c[/tex] in your solution.
im trying to figure out what you mean here, ill keep on it.

RoyalCat said:
Is the expected answer you posted the textbook's answer? I'm asking since it makes absolutely no sense.. If we are to understand that [tex]x_0=x_c[/tex], then [tex]f(x_0 \frac{b}{k})=f(1)[/tex] and that doesn't make sense dimensionally. But let's assume that the meaning was just 1, then all you get is just the linear frequency.

Where did you get this question..?
Introductory Classical Mechanics,
with Problems and Solutions
David Morin

in the appendix pg XIV-6
 
  • #4
Mechdude said:
yes your right,

i hope i did not refer to SHM,


im trying to figure out what you mean here, ill keep on it.


Introductory Classical Mechanics,
with Problems and Solutions
David Morin

in the appendix pg XIV-6

I was hoping it was a book I had access to. ^^; Could you please scan in the question, or write it in full? Since as it stands, it doesn't make sense. Especially the answer from the answer sheet which would be dimensionally inconsistent.

The force is a function of distance, you can't plug in something which has the dimensions of a pure number ([tex]x_0\frac{b}{k}[/tex] would be a pure number) and get something that makes sense.
 
  • #5
here is a screenshot of the book , hope its clearer than my explanation
 

Attachments

  • dimensianalanalysiscpt.png
    dimensianalanalysiscpt.png
    51.3 KB · Views: 454
  • #6
Oh, I think I understand now. The lowercase [tex]f[/tex] just stands for some function whose argument is a pure number.

This stems from two assumptions. The first is that there a dependence of the frequency on the parameter [tex]b[/tex].

This is problematic since standard dimensional analysis (Assuming [tex]\omega = c\cdot m^{\alpha}\cdot k^{\beta} \cdot b^{\gamma}[/tex]) shows that there cannot be a dependence on [tex]b[/tex] because of the length dimension it introduces.

We solve this problem by assuming a dependence on the amplitude of the motion as well. This way we can negate the dimension of length by saying that we have a dependence on [tex]x_0 b[/tex]

We do this by saying that the new frequency is the same as the old, times some pure number, [tex]f(x_0\frac{b}{k})[/tex] which is a function of a pure number just as well.
 

What is dimensional analysis?

Dimensional analysis is a mathematical method used to convert units of measurement from one system to another. It involves using conversion factors and canceling out units to ensure that the final result has the desired units.

Why is dimensional analysis important in science?

Dimensional analysis is important in science because it helps to ensure accurate and consistent measurements. It also allows scientists to compare and combine data from different experiments and studies, even if they use different units of measurement.

How do you set up a dimensional analysis problem?

To set up a dimensional analysis problem, you need to start with the known quantity and its units. Then, you need to determine the conversion factors needed to convert those units into the desired units. Finally, you can set up a proportion and solve for the unknown quantity.

What is a conversion factor in dimensional analysis?

A conversion factor is a ratio that relates two different units of measurement. It is used in dimensional analysis to convert between units by canceling out unwanted units and leaving the desired units in the final answer.

Can dimensional analysis be used for any type of unit conversion?

Yes, dimensional analysis can be used for any type of unit conversion as long as the two units are related by a conversion factor. However, it is important to be careful and ensure that the units are being converted correctly, as using the wrong conversion factor can lead to incorrect results.

Suggested for: Dimensional analysis problem

Replies
4
Views
750
Replies
8
Views
1K
Replies
15
Views
949
Replies
24
Views
1K
Replies
15
Views
870
Replies
2
Views
604
Replies
47
Views
510
Back
Top