Handling two SHMs in different directions

[palaphys]

Homework Statement

Two SHMs given by the equations x1=A1sin wt and x2=A2 sin(wt+c) operate at an angle theta with each other. Find the resultant SHM. {Question asked only for amplitude but I want to know about the SHM itself.}

Relevant Equations

X=Asinwt

So when I attempted this question I was kind of stumped, as I have dealt only with SHMs in the same direction, which I can just add up.

I know that this is possible because when we represent the SHMs as PHASORS, they rotate with some constant angular speed and can be treated as vectors. How do I deal with this one though? Is it even possible to find the resultant SHM?

I also read about Lissajous figures which was a particular case when the angle was 90 in between the SHMs. In that case it was easy to come up with an equation for a tilted ellipse, which changes based on phase difference.

Please help, I am confused whether or not to consider SHMs as vectors for finding the resultant.
Do I just multiply a term of e^i(theta) to the term A2 e^i(wt+c) ? But that seems to change the phase of the phasor and seems incorrect to me

 

[pasmith]

Do you not have <br /> \mathbf{x}(t) = x_1(t) \mathbf{e}_1 + x_2(t)(\cos \theta \mathbf{e}_1 + \sin \theta \mathbf{e}_2)? Then you can write the \mathbf{e}_1 component as R_1 \sin (\omega t + c_1) where R_1 and c_1 are functions of A_1, A_2, c, \theta.

 

[palaphys]

Do you not have <br /> \mathbf{x}(t) = x_1(t) \mathbf{e}_1 + x_2(t)(\cos \theta \mathbf{e}_1 + \sin \theta \mathbf{e}_2)? Then you can write the \mathbf{e}_1 component as R_1 \sin (\omega t + c_1) where R_1 and c_1 are functions of A_1, A_2, c, \theta.

What is e1?

 

[WWGD]

What is e1?

A basis vector.

 

[palaphys]

A basis vector.

How is it possible to write this Simple harmonic motin as a vector?

 

[Steve4Physics]

Homework Statement: Two SHMs given by the equations x1=A1sin wt and x2=A2 sin(wt+c) operate at an angle theta with each other.

This is a bit ambiguous.

a) Are 'c' and 'angle theta' meant to be the same thing (the phase difference) with both SHMs along the same x-axis?

b) Or are the two SHMs along different axes with angle $\theta$ between the axes?

Also, best to use LaTeX and write, for example, $x_1 = A_1 \sin(\omega t)$. See LateX-guide link at bottom-left of edit-window (or use https://www.physicsforums.com/help/latexhelp/).

I also read about Lissajous figures

A Lissajous figure is produced when one SHM is along the x-axis and the other is along the y-axis. It's a curve in the xy plane.

I'm guessing that what you want to do is a), above. This is equivalant to adding 2 phasors of different amplitudes with a phase-difference of $\theta$. If that's what you mean, try this: https://scipp.ucsc.edu/~haber/ph5B/addsine.pdf

Edit - typo' fixed.

 

[palaphys]

This is a bit ambiguous.

a) Are 'c' and 'angle theta' meant to be the same thing (the phase difference) with both SHMs along the same x-axis?

b) Or are the two SHMs along different axes with angle $\theta$ between the axes?

Also, best to use LaTeX and write, for example, $x_1 = A_1 \sin(\omega t)$. See LateX-guide link at bottom-left of edit-window (or use https://www.physicsforums.com/help/latexhelp/).


A Lissajous figure is produced when one SHM is along the x-axis and the other is along the y-axis. It a curve in the xy plane.

I'm guessing that what you want to do is a), above. This is equivalant to adding 2 phasors of different amplitudes with a phase-difference of $\theta$. If that's what you mean, try this: https://scipp.ucsc.edu/~haber/ph5B/addsine.pdf

A) c is the phase difference, b is the situation we have here

 

[Steve4Physics]

A) c is the phase difference, b is the situation we have here

OK, you are combining 2 linear SHMs in different directions. For example, when the angle between the two directions is $\theta = 90^{\circ}$ we get Lissajous figures.

In Post #1, the question is "Find the resultant SHM." The problem is that the resultant motion is not itself linear SHM; it can't be expressed as a single equation of the form $x(t) =A \sin(\omega t + \phi)$ because the motion is in 2D.

At the risk of overlapping with what @pasmith has said...

The resultant position-vector is $\mathbf{x}(t) = x_1(t) \mathbf{e_1} + x_2(t) \mathbf{e_2}$ where $\mathbf{e_1}$ and $\mathbf{e_2}$ are the unit vectors for each SHM’s line of motion, with angle $\theta$ between them.

So the resultant motion (in 2D) is given by:
$\mathbf{x}(t) = A_1 \sin(\omega t) \mathbf{e_1} + A_2 \sin(\omega t+c) \mathbf{e}_2$

$\theta$ would appear if you wanted to use different unit vectors (a different basis). E.g. suppose you know $\mathbf{e_1}$ and $\mathbf{e_2}$ in terms if Cartesian coordinates and you want to express the position-vector in terms of Cartesian coordinates, $(x(t), y(t))$, then that’s where $\theta$ appears.

 

[palaphys]

So the resultant motion (in 2D) is given by:
$\mathbf{x}(t) = A_1 \sin(\omega t) \mathbf{e_1} + A_2 \sin(\omega t+c) \mathbf{e}_2$

$\theta$ would appear if you wanted to use different unit vectors (a different basis). E.g. suppose you know $\mathbf{e_1}$ and $\mathbf{e_2}$ in terms if Cartesian coordinates and you want to express the position-vector in terms of Cartesian coordinates, $(x(t), y(t))$, then that’s where $\theta$ appears.

okay, but could I decompose the two SHMs into x components and y components? Like in the y direction there is an $$ A_2 \sin(\omega t + \psi) \cos\theta$$ and along the x axis we have $$A_1 \sin(\omega t) + A_2 \sin(\omega t + \psi) \sin\theta$$

So can I just say that y= $$ A_2 \sin(\omega t + \psi) \cos\theta$$
x=$$A_1 \sin(\omega t) + A_2 \sin(\omega t + \psi) \sin\theta$$
And now I establish a relation with x and y to get the path of the object?
What about the resultant amplitude? What would that be?
Please correct me if I'm wrong.

 

[kuruman]

$\theta$ would appear if you wanted to use different unit vectors (a different basis). E.g. suppose you know $\mathbf{e_1}$ and $\mathbf{e_2}$ in terms if Cartesian coordinates and you want to express the position-vector in terms of Cartesian coordinates, $(x(t), y(t))$, then that’s where $\theta$ appears.

I would say that it is given that $$\cos\!\theta=\mathbf{e_2}\cdot \mathbf{e_1}.$$

 

[kuruman]

okay, but could I decompose the two SHMs into x components and y components? Like in the y direction there is an $$ A_2 \sin(\omega t + \psi) \cos\theta$$ and along the x axis we have $$A_1 \sin(\omega t) + A_2 \sin(\omega t + \psi) \sin\theta$$

So can I just say that y= $$ A_2 \sin(\omega t + \psi) \cos\theta$$
x=$$A_1 \sin(\omega t) + A_2 \sin(\omega t + \psi) \sin\theta$$
And now I establish a relation with x and y to get the path of the object?
What about the resultant amplitude? What would that be?
Please correct me if I'm wrong.

First, you need to write complete vector equations, not just bits and pieces like $$A_2 \sin(\omega t + \psi) \cos\theta$$ that express the position of each oscillator as a function of time. Then add the two as vectors to find the resultant.

 

[palaphys]

First, you need to write complete vector equations, not just bits and pieces like $$A_2 \sin(\omega t + \psi) \cos\theta$$ that express the position of each oscillator as a function of time. Then add the two as vectors to find the resultant.

But how isnt what I wrote just the y component? Also how is an SHM a vector? I dont seem to understand.

 

[Steve4Physics]

okay, but could I decompose the two SHMs into x components and y components? Like in the y direction there is an $$ A_2 \sin(\omega t + \psi) \cos\theta$$ and along the x axis we have $$A_1 \sin(\omega t) + A_2 \sin(\omega t + \psi) \sin\theta$$

So can I just say that y= $$ A_2 \sin(\omega t + \psi) \cos\theta$$
x=$$A_1 \sin(\omega t) + A_2 \sin(\omega t + \psi) \sin\theta$$
And now I establish a relation with x and y to get the path of the object?
What about the resultant amplitude? What would that be?
Please correct me if I'm wrong.

That approach looks valid in principle. A couple of comments...

You are defining SHM-1 to be along the x-axis. If SHM-2 is along a line at angle $\theta$ counter-clockwise to the x-axis (standard convention) I think you have $\sin \theta$ and $\cos \theta$ mixed-up. So you should have (using $c$ rather than $\psi$):

$x(t) = A_1 \sin(\omega t) + A_2 \sin(\omega t + c) \cos\theta$
$y(t) = A_2 \sin(\omega t + c) \sin\theta$

I'm guessing you want the equation of the trajectory - probably a tilted ellipse. This would require eliminating $t$ from the above equations leaving an equation relating $x, y$ and constants (looks messy!).

There is no 'amplitude' in the usual sense because the motion is a trajectory in 2D - it is not linear SHM. Maybe you want the maximum magnitude of the distance from the origin. This is the maximum value of $\sqrt {x^2+y^2}$ and (for an elliptical trajectory) is the length of the semi-major axis.

 

[palaphys]

That approach looks valid in principle. A couple of comments...

You are defining SHM-1 to be along the x-axis. If SHM-2 is along a line at angle $\theta$ counter-clockwise to the x-axis (standard convention) I think you have $\sin \theta$ and $\cos \theta$ mixed-up. So you should have (using $c$ rather than $\psi$):

$x(t) = A_1 \sin(\omega t) + A_2 \sin(\omega t + c) \cos\theta$
$y(t) = A_2 \sin(\omega t + c) \sin\theta$

I'm guessing you want the equation of the trajectory - probably a tilted ellipse. This would require eliminating $t$ from the above equations leaving an equation relating $x, y$ and constants (looks messy!).

There is no 'amplitude' in the usual sense because the motion is a trajectory in 2D - it is not linear SHM. Maybe you want the maximum magnitude of the distance from the origin. This is the maximum value of $\sqrt {x^2+y^2}$ and (for an elliptical trajectory) is the length of the semi-major axis.

Yes,thanks for correcting. I indeed mixed the sin and cos terms. Okay I get that the trajectory is a tilted ellipse. But one thing that's bugging me is that how are we able to vectorially add SHMs? It is very strange. Is there a reason behind it?

 

[kuruman]

I'm guessing you want the equation of the trajectory - probably a tilted ellipse.

The problem asks "Find the resultant SHM". In my opinion this is ambiguous. Each mass undergoes SHM along a straight line separately. So when you say the "trajectory", it's the trajectory of what? Where does this ellipse come from? In Lissajous figures there is only one mass that follows the trajectory.

The only interpretation of "Find the resultant SHM" that I can think of is to find the scalar distance between the two masses as a function of time. It's a straight line of variable length $L$ that has period $T=2\pi/\omega.$

I would write
$\mathbf {x}_1=A_1\sin\omega t~\mathbf e_1$
$\mathbf {x}_2=A_2\sin(\omega t+c)(\cos\!\theta~\mathbf e_1+\sin\!\theta~\mathbf e_2)$

and then find an expression for $L(t)=|\mathbf {x}_2-\mathbf {x}_1|^{1/2}.$

 

[palaphys]

The problem asks "Find the resultant SHM". In my opinion this is ambiguous. Each mass undergoes SHM along a straight line separately. So when you say the "trajectory", it's the trajectory of what? Where does this ellipse come from? In Lissajous figures there is only one mass that follows the trajectory.

The only interpretation of "Find the resultant SHM" that I can think of is to find the scalar distance between the two masses as a function of time. It's a straight line of variable length $L$ that has period $T=2\pi/\omega.$

I would write
$\mathbf {x}_1=A_1\sin\omega t~\mathbf e_1$
$\mathbf {x}_2=A_2\sin(\omega t+c)(\cos\!\theta~\mathbf e_1+\sin\!\theta~\mathbf e_2)$

and then find an expression for $L(t)=|\mathbf {x}_2-\mathbf {x}_1|^{1/2}.$

There is only one mass here. By two SHMs I think the question means two restoring forces which can cause a simple harmonic motion

 

[Steve4Physics]

The problem asks "Find the resultant SHM". In my opinion this is ambiguous.

Yes, the setup and question(s) aren’t entirely clear.

I interpreted the system to be a single particle moving in a 2D plane as a result of simultaneously performing two linear SHMs along different lines (c.f. Lissajous figures).

It would help if the OP could confirm this or not.

EDIT. @palaphys confirmed my interpretation in Post #16.

 

[kuruman]

There is only one mass here. By two SHMs I think the question means two restoring forces which can cause a simple harmonic motion

If you think that there is only mass, so be it. You say you want to know about the SHM itself. If by this you mean the position of the single mass as a function of time, it will be given (as stated in post #2 by @pasmith ) by
$\mathbf r=A_1\sin\omega t~\mathbf {\hat x}+A_2\sin(\omega t+c)(\cos\!\theta~\mathbf {\hat x}+\sin\!\theta~\mathbf {\hat y})~$
in a Cartesian coordinate system where the first SHM is along the $x-$axis and the second SHM at angle $\theta$ with respect to the $x-$axis.

 

[WWGD]

How is it possible to write this Simple harmonic motin as a vector?

Well, it takes place in two dimensions; in the Plane, in $\mathbb R^2$.