Is Hooke's law really not working, or is it me being dummy?

In summary: If you have a system made of a spring and a mass on a frictionless surface, just like in the first picture. If you pull the mass away from its equilibrium position and reach X1, the restoration force of the spring at X1 will be F1. If you keep pulling it and reach the X2 point, you will have an F2. If X2=2.X1 then F2 should also be two times bigger than F1 based on the hook's low. correct? So the graph will be something like the second picture.The graph is a sinusoidal wave.
  • #1
Ara wolf
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TL;DR Summary
I'm wondering why do we have a sinusoidal wave instead of a triangular wave when we describe a simple oscillation.
Hello everyone,

I'm a new member, and you might see me around from now on. I'm now on a path to understanding the mathematics behind a complicated mechanical machine. My knowledge is basically what I learned during my school days and also during university courses, and for me, it was mostly based on memorizing the formulas and doing homework! Now I'm trying to understand the real fundamentals. Before I start, I should apologize for my poor English, Please just bear with me. So here is my question:

Imagine you have a system made of a spring and a mass on a frictionless surface, just like in the first picture. If you pull the mass away from its equilibrium position and reach X1, the restoration force of the spring at X1 will be F1. If you keep pulling it and reach the X2 point, you will have an F2. If X2=2.X1 then F2 should also be two times bigger than F1 based on the hook's low. correct? So the graph will be something like the second picture. Now imagine that after pulling it to X2 you let the object go, the spring will pull it, it goes all the way to the left, reaches -X2, and reaches -X2, and then comes back to X2. Since we have no friction, there will be no waste of energy, and the restoration force for any X should be the same any time it passes that point. it means if I want to stop the object with my hand at X1, I should do so by a applying a force as big as the one I applied the first time that I pulled the spring so far to X1 which is also as big as the restoration force of F1 but in the opposite direction. So it should exactly track the same graph of the second picture each time. That means the graph should be a triangular wave, like in the 3rd picture. And since the force is F=ma and your mass in this example is constant, the graph for acceleration should also look similar, but this movement is a simple harmonic oscillation, and the graph for acceleration in a simple harmonic system is shown as a sinusoidal wave! So am I losing some points? Is Hook's law not working, or is there a problem with my logic?
slide_12.jpgimages (1).pnggraph.jpg
 
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  • #2
Welcome to PF. :smile:

In your schooling so far, have you learned any calculus, specifically Differential Equations? It's a lot more intuitive to understand a spring-mass sinusoidal oscillation if you're comfortable with the DE that describes the motion.
 
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  • #3
Let's start with something simpler. Your third drawing, labeled force vs. displacement, does not express Hooke's Law.
 
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  • #4
Ara wolf said:
TL;DR Summary: I'm wondering why do we have a sinusoidal wave instead of a triangular wave when we describe a simple oscillation.

That means the graph should be a triangular wave, like in the 3rd picture.
The 3rd picture is just not a plot of oscillation at all. Oscillation would have displacement on the vertical axis and time on the horizontal axis.
 
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  • #6
I think the OP's calculation is that if ##F = kx##, then ##a = \frac k m x## (okay so far), but then (and this is the problem) ##x = \frac k m t##.

The last equation, however, represents constant velocity, with ##v = \frac k m## and zero acceleration. And not the varying acceleration we have in a spring system.
 
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  • #7
@Ara wolf in addition to my points above, note also:

You have force in the same direction as displacement. You should have ##F = - kx##.

You have a mysterious infinite acceleration at the turning points, where the force mysteriously changes direction.
 
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  • #8
Ara wolf said:
the graph for acceleration should also look similar,
What graph for acceleration? Acceleration vs. time or acceleration vs. displacement? You seem to ignore what is on the x-axis, as your confusing 3rd plot also shows.
 
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  • #9
That's the result of teaching mechanics in a "calculus-free" way. Without calculus you can't understand most motions. Here it's pretty simple, and that's what makes the harmonic oscillator so important. Also it's a good approximation in almost all cases, when you have a bound motion, i.e., the motion of a particle in a potential pot around the equilibrium position. So it's well worth to treat this example carfully and with the right tools, i.e., calculus, and it's very simple calculus.

The equation of motion reads
$$ma=m \ddot{x}=-k x.$$
or
$$\ddot{x} = -\frac{k}{m} x.$$
That's a differential equation, and usually it's not easy to solve such equations, but here indeed it is simple, because we just have to think, which functions are their own 2nd derivatives up to (negative) factor. These are obviously cos and sin functions. Now the equation is also linear, i.e., if you have two solutions any linear combination of those are also solutions. That implies that the ansatz
$$x(t)=A \cos(\omega t) + B \sin(\omega t)$$
should work. Indeed, taking the derivatives gives
$$\dot{x}(t)=-A \omega \sin(\omega t) + B \omega \cos(\omega t)$$
and
$$\ddot{x}(t)=-A\omega^2 \cos(\omega t) - B \omega^2 \sin(\omega t)=-\omega^2 x(t).$$
So, as expected, the ansatz leads to a solution: Plugging it into the equation of motion, gives
$$\omega^2=\frac{k}{m},$$
and we already get the complete set of solutions by choosing the positive frequency
$$\omega=\sqrt{\frac{k}{m}}.$$
The negative solution, leads to the same set of solutions, since the two "integration constants", ##A## and ##B##, can be chosen arbitrarily.

They are determined by giving initial conditions, i.e., ##x(0)=x_0## and ##\dot{x}(0)=v_0##. Using our ansatz gives
$$x(0)=A=x_0, \quad \dot{x}(0)=B \omega = v_0 \; \Rightarrow\; B=\frac{v_0}{\omega}.$$
So the solution of the initial-value problem finally is
$$x(t)=x_0 \cos(\omega t) + \frac{v_0}{\omega} \sin(\omega t) \quad \text{with} \quad \omega=\sqrt{\frac{k}{m}}.$$
 
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  • #10
Ara wolf said:
... So it should exactly track the same graph of the second picture each time. That means the graph should be a triangular wave, like in the 3rd picture. ..... is there a problem with my logic?
Welcome!

The problem that I see with your logic regarding the third picture is that the displacement is restricted to happen between the points X2 and -X2.

Please, see:
https://www.physicsclassroom.com/calcpad/Simple-Harmonic-Motion/Equation-Overview

https://courses.lumenlearning.com/s...n, the,opposite direction of the displacement.
 
  • #11
I think the OP is almost correct, but just a bit confused about the triangular graph. In that graph the horizontal axis is the displacement. But notice that there are two x-positions where the value is supposed to be X1. Also, there is a value of -X2 to the *right* of X2. This should be on the left side of course.

I think the OP is mixing up time and position in this graph, he plots the horizontal axis as if it was time, with a the labels showing position. But in that case the assumption of linearity between what is actually the time on the horizontal axis and force is wrong, this should indeed be a sinusoidal curve.

If you would actually make a plot with the position against force the OP would be correct, that is, if -X2 would be positioned left of X1, if there would be only a single X1, X2 label etc. In this case force against position is indeed a triangular curve. And also the acceleration plotted against the position is indeed a triangular curve. But in that case the lines of each oscillation overlap.
 
  • #12
Arjan82 said:
In that graph the horizontal axis is the displacement.
distance, not displacement.
force vs displacement is just a diagonal [edit: through the origin] : ##F=-kd##.

What the OP (and everybody else) wants is force vs time [edit: displacement vs time ; geez I really gotta start proofreading].
 
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  • #13
hmmm27 said:
distance, not displacement.

1688414319960.png


I was reacting to what the OP had written down....

hmmm27 said:
force vs displacement is just a diagonal : ##F=-kd##.

That was exactly my point indeed....

hmmm27 said:
What the OP (and everybody else) wants is force vs time.

What you think the OP wants is not what he has written down. I can only react to what he has written. There he clearly states that he is surprised that the force against 'displacement' (indeed position) is linear. Which is what I reacted to.
 
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  • #14
Arjan82 said:
What you think the OP wants is not what he has written down.
My bad, I meant "displacement vs time", since the question was about the sine wave : edited.
 

Related to Is Hooke's law really not working, or is it me being dummy?

What is Hooke's Law?

Hooke's Law states that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it is expressed as F = -kx, where F is the force applied, k is the spring constant, and x is the displacement from the equilibrium position.

In what situations does Hooke's Law not apply?

Hooke's Law is only valid within the elastic limit of the material. If the material is stretched or compressed beyond this limit, it will not return to its original shape, and Hooke's Law will no longer apply. This is known as plastic deformation. Additionally, Hooke's Law does not apply to materials that do not exhibit linear elastic behavior, such as rubber or biological tissues.

How do I know if I'm applying Hooke's Law correctly?

To ensure you're applying Hooke's Law correctly, make sure that the material you're working with is within its elastic limit and that the relationship between force and displacement is linear. Also, ensure your measurements are accurate and that you're using the correct value for the spring constant (k). If these conditions are met and your results still seem off, there could be experimental errors or other forces at play.

What could cause deviations from Hooke's Law in an experiment?

Deviations from Hooke's Law in an experiment can be caused by several factors, including reaching the material's elastic limit, inaccuracies in measuring force or displacement, imperfections in the material, temperature changes, or additional forces acting on the system. Ensuring precise measurements and controlling experimental conditions can help minimize these deviations.

Is it possible for a spring to have a non-linear response even if it's within the elastic limit?

Yes, it is possible for some materials to exhibit non-linear elastic behavior even within their elastic limits. This means that the relationship between force and displacement is not a straight line. Such materials require more complex models to describe their behavior. If you're working with a material that doesn't follow a linear response, Hooke's Law may not be applicable, and you may need to use a different model to describe its elasticity.

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