Interpreting Simple Harmonic Motion Graphs

[Jimmy87]

Homework Statement

Explain the shape of the velocity-displacement and acceleration-displacement graphs for an object undergoing simple harmonic motion. The graph is attached to this thread

Homework Equations

v = wsqrt(A^2-x^2) where w = angular frequency, A = amplitude and x = displacement

a = - w^2x

The Attempt at a Solution

I get the acceleration-displacement graph. The acceleration is proportional to the displacement which means you will get a straight line (with a negative slope because of the minus sign). I can't figure out why the velocity-displacement graph is circular?

 

[mfb]

I get the acceleration-displacement graph. The acceleration is proportional to the displacement which means you will get a straight line (with a negative slope because of the minus sign). I can't figure out why the velocity-displacement graph is circular?

What is the velocity at the points of maximal displacement? What is the displacement at the point of maximal velocity (positive and negative)? How do they change in between?

 

[Jimmy87]

What is the velocity at the points of maximal displacement? What is the displacement at the point of maximal velocity (positive and negative)? How do they change in between?

Velocity at maximum displacement is zero and at minimum displacement is maximum. I see that on the graph but why the curved shape between these two points and why does the whole thing trace out a circle?

 

[mfb]

If you know the formula for displacement over time, you can calculate its first derivative (or look it up) to get the velocity.
If you square both and add them, you get an ellipse - or a circle as special case if you choose the axes in the right way.

This is not by accident, it is related to energy conservation: both velocity squared and distance squared correspond to components of the total energy, so points of equal distance to the origin correspond to equal energy.

 

[BvU]

Do you have an expression for x as a function of t ?
What does that yield when substituted in v = wsqrt(A^2-x^2) ?

Personally, I like the animation in this Wikipedia lemma.

 

[Jimmy87]

Do you have an expression for x as a function of t ?
What does that yield when substituted in v = wsqrt(A^2-x^2) ?

Personally, I like the animation in this Wikipedia lemma.

So, x = A cos(wt) so you would get v = wsqrt(A^2-A^2cos^2w^2t^2). How can I use this to explain the shape of the velocity curves?

 

[Jimmy87]

I have been searching online for some answers and may have found one. Is the equation given the equation for a half circle and since the velocity can have two values (positive and negative) you get a full circle, is that right? See example 4 in this link (http://www.analyzemath.com/Graphing/graphing_square_root_func.html) or have I got it completely wrong. We haven't done any circle maths in class so I didn't think it would be this complicated...

 

[Jimmy87]

Does anyone have any answers please?

 

[BvU]

I see, no circle math. But then I would like to know where your v = wsqrt(A^2-x^2) came from (because that seems to be the hangup now). It is also wrong, so we need to clarify and fix that.

v has one single value at any time, just like x and a. But its value range is from -$\omega$A to +$\omega$A.

 

[BvU]

Yes. So I have to correct my silent approval of

x = A cos(wt) so you would get v = wsqrt(A^2-A^2cos^2w^2t^2).

Because that expression yields only non-negative v, and it is clear that the motion involves negative v half the time!

There is the relationship $v^2 = \omega^2 (A^2 - x^2)$ which is true at all times and does help to draw the graph. It just doesn't help finding v at a given x, because for every x there are two v that satisfy the relationship: same |v| but opposite sign.

Again, I really like the wikipedia picture (which is your picture turned 1/4 turn).

 

[Jimmy87]

This

I have been searching online for some answers and may have found one. Is the equation given the equation for a half circle and since the velocity can have two values (positive and negative) you get a full circle, is that right? See example 4 in this link (http://www.analyzemath.com/Graphing/graphing_square_root_func.html) or have I got it completely wrong. We haven't done any circle maths in class so I didn't think it would be this complicated...

This link has the equation in it before it has been square rooted (http://www.mathsrevision.net/advanced-level-maths-revision/mechanics/simple-harmonic-motion). If you look under further equations it is listed there.

 

[Jimmy87]

Sorry only just seen your second reply. I still don't understand the shape of the graph though? How does that equation give rise to the graph I originally attached?

 

[Jimmy87]

Please could someone kindly explain this graph, I still don't get it! The equation and animation posted by BvU shows a circular shape. What does that equation say is the relationship between displacement and velocity? Is it that the velocity is proportional to the square root of some function of x squared? If you plot any graph using an equation you should be able to predict the outcome using the equation and I don't see how that equation should give a circle/curve?

v2=ω2(A2−x2) - this is the equation I mean which gives rise to this circular shaped graph if you plot velocity against dispalcement

 

[mfb]

Is it that the velocity is proportional to the square root of some function of x squared?

Yes, but that description is very broad. "Some function" is a specific function.
If you did not have functions for circles yet, this is hard to understand. Did you calculate distances on a plane before? Or the length of a side in a rectangular triangle? Those formulas are useful here.

 

[nasu]

It's not a circular shape but an ellipse. So you need to know the equation of the ellipse in Cartesian coordinates. Then you can show that V and x satisfy the equation of the ellipse in the phase space. It's quite simple.

 

[Jimmy87]

Yes, but that description is very broad. "Some function" is a specific function.
If you did not have functions for circles yet, this is hard to understand. Did you calculate distances on a plane before? Or the length of a side in a rectangular triangle? Those formulas are useful here.

Thanks. We haven't done anything on functions of circles. So does that equation describe the equation of a circle and hence why the graph is circular? I have attached a small extract from my textbook. At the end of the first main paragraph it says that this shape for the velocity displacement graph is suprising and gives some vague guidance on how to show why the shape of the graph is like that. Does the guidance it give (dealing with cos^2theta and sine^2theta) make any sense to anyone?

 

[Jimmy87]

It's not a circular shape but an ellipse. So you need to know the equation of the ellipse in Cartesian coordinates. Then you can show that V and x satisfy the equation of the ellipse in the phase space. It's quite simple.

Is it simple for a pre-college physics student? Could you point me in the direction of source that links this to the equation of an ellipse? I always thought SHM is all to do with circular motion and didn't involve ellipses?

 

[Jimmy87]

Sorry this should be clearer, think the last attachment was too small to see

 

[nasu]

Well, how can I guess your level? It does not seem you mentioned it.:)
Here you can find the equation of an ellipse.
http://en.wikipedia.org/wiki/Ellipse (see under "Equations")
The one you need here is
(x/a)^+(y/b)^2=1
Here a and b are the semi-axes of the ellipse.

You show an elliptical shape in the problem so if you want to solve it you have to deal with ellipses, don't you? Or give up until you learn about ellipses. You cannot pretend is a circle just because you did not cover this yet. :)

Hint. In your case, try to calculate (x/A)^2+(v/wA)^2. Which is what they suggest at the end of the paragraph in your picture.

 

[Jimmy87]

Well, how can I guess your level? It does not seem you mentioned it.:)
Here you can find the equation of an ellipse.
http://en.wikipedia.org/wiki/Ellipse (see under "Equations")
The one you need here is
(x/a)^+(y/b)^2=1
Here a and b are the semi-axes of the ellipse.

You show an elliptical shape in the problem so if you want to solve it you have to deal with ellipses, don't you? Or give up until you learn about ellipses. You cannot pretend is a circle just because you did not cover this yet. :)

Hint. In your case, try to calculate (x/A)^2+(v/wA)^2. Which is what they suggest at the end of the paragraph in your picture.

Thanks I will try and digest all of this. So when you get taught that simple harmonic is modeled using circular motion this isn't necessarily true because if you are dealing with velocity and displacement you are technically dealing with ellipses? We were shown a ping pong ball going around in a circle which casts a 1 dimensional shadow which undergoes simple harmonic motion and we applied the equations of circular math to SHM.

 

[nasu]

No, your conclusion is not right. Yes, indeed the circular motion can be modeled as two harmonic motions along the x and y axes.
Or a SHM motion can be modeled as the projection of the circular motion on one of the axes. So in this sense there is a connection between SHM and circular motion.

But it does not follow than anything you do or plot about SHM will be a circle, does it?
What if you plot x versus t? Will this be a circle or a sine wave? Are you surprised that it is a sine wave and not a circle?
In this problem you have a completely different thing, a plot of v versus x. I don't see how the above analogy between SHM and circular motion will grant that this plot will be a circle.

 

[Jimmy87]

No, your conclusion is not right. Yes, indeed the circular motion can be modeled as two harmonic motions along the x and y axes.
Or a SHM motion can be modeled as the projection of the circular motion on one of the axes. So in this sense there is a connection between SHM and circular motion.

But it does not follow than anything you do or plot about SHM will be a circle, does it?
What if you plot x versus t? Will this be a circle or a sine wave? Are you surprised that it is a sine wave and not a circle?
In this problem you have a completely different thing, a plot of v versus x. I don't see how the above analogy between SHM and circular motion will grant that this plot will be a circle.

I see your point, I think I was assuming it would be. I thought that a sine wave was kind of a circle since it repeats itself in cycles (circles). Is it bad to think of the two halves of a sine wave as being a circle if you put them together?

 

[mfb]

Is it bad to think of the two halves of a sine wave as being a circle if you put them together?

They are not.

 

[nasu]

I see your point, I think I was assuming it would be. I thought that a sine wave was kind of a circle since it repeats itself in cycles (circles). Is it bad to think of the two halves of a sine wave as being a circle if you put them together?

Yes, it is bad because they are not. Not any closed loop is a circle, is it?

 

[ehild]

We were shown a ping pong ball going around in a circle which casts a 1 dimensional shadow which undergoes simple harmonic motion and we applied the equations of circular math to SHM.

When modelling SHM as the 1 dimensional shadow of the ball moving around in a circle, you got that the displacement of the shadow from equilibrium is x= A sin (wt). It was also derived that the projection of the velocity vector is Aw cos (wt), and that of the acceleration is -Aw²sin(wt). Check your lecture notes.

Now you have two equations, x=Asin(wt) and v=Aw cos(wt).
They are the parametric equations describing a line in the (x,v ) plane. (You take v as y)

You can write the equations as x/A= cos(wt) and v/(wA) = cos (wt)
Square both equations and add them together. You get
\left(\frac{x}{A}\right)^2+\left(\frac{v}{wA}\right)^2=\sin^2(wt)+\cos^2(wt)=1
or
\frac{x^2}{A^2}+\frac{v^2}{(wA)^2}=1
But that is the equation of an ellipse with axes A and wA in the (x,v) plane. If you have not covered ellipse yet, see that: http://www.mathwarehouse.com/ellipse/equation-of-ellipse.php