# Homework Help: Proof of Standing Waves

1. Jun 8, 2014

### alingy1

Whenever we prove standing waves in books, they define two waves:
D1(x,t)=asin(kx-wt)
D2(x,t)=asin(kx+wt)

Why don't we ever put phase constants?
What if the phase constants for each is different? Will there be a standing wave?

2. Jun 9, 2014

### Simon Bridge

Do you know what the phase constant means?

Why not put phase constant in and see if they make a difference.

(Nobody say: it is a good exercise and alingy1 will learn a lot.)

3. Jun 9, 2014

### alingy1

What happens from here?

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4. Jun 9, 2014

### alingy1

Sometimes, I wish I had studied trigonometry harder back in high school! *sigh*

5. Jun 9, 2014

### Simon Bridge

OK so, from your attachment, you did: $$D_R=\sin(kx-\omega t+\phi)\\ D_L=\sin(kx+\omega t)\\ D_R+D_L = \sin(kx-\omega t+\phi) + \sin(kx+\omega t)\\ \qquad = \sin kx \cos(\omega t+\phi) - \cos kx\sin(\omega t + \phi) + \sin kx \cos \omega t + \cos kx\sin \omega t$$(... I left off the constant amplitude "a" to save typing.)

Have you forgotten a minus sign when you applied the angle sum/difference formula?
vis: $\sin(kx-\omega t +\phi) = \sin\big(kx-(\omega t - \phi)\big) = \sin(A-B)$
http://en.wikipedia.org/wiki/List_of_trigonometric_identities#Angle_sum_and_difference_identities

... the approach from here should parallel the one you normally see in the regular derivation.
You have to keep going through trig identities and group like terms and so on.

It may help you if you apply the angle sum formula on the $\omega t+\phi$ functions, it should give you a bunch of terms you are used to, or try grouping terms in $kx$ ... but it is usually easier to handle sine functions by using the Euler equation right from the start.$$D_R+D_L= \frac{1}{2i}\left[ e^{i(kx-\omega t +\phi)} - e^{-i(kx-\omega t +\phi)} + e^{i(kx+\omega t)} - e^{-i(kx+\omega t)}\right]$$... it's the same approach without having to use trig identities.

But it is interesting that you have intuitively grouped the phase angle with the frequency-time part. In the regular derivation - where does the ωt part end up?

Definitely group terms in kx and another identity should occur to you.

Last edited: Jun 9, 2014
6. Jun 12, 2014

### alingy1

I am plain lost. I intuitively did it because it was one of the two possibilities of grouping.
I do not know which identities I should use. My math was always very weak.

7. Jun 12, 2014

### Simon Bridge

$$D_R=\sin(kx-\omega t+\phi)\\ D_L=\sin(kx+\omega t)\\ D_R+D_L = \sin(kx-\omega t+\phi) + \sin(kx+\omega t)\\ \qquad = \sin kx \cos(\omega t+\phi) - \cos kx\sin(\omega t + \phi) + \sin kx \cos \omega t + \cos kx\sin \omega t\\ \qquad = \big(\cos(\omega t+\phi)+\cos \omega t \big)\sin kx + \big( \sin \omega t-\sin(\omega t + \phi)\big)\cos kx\qquad$$

sum-to-product identities on the terms in brackets:
http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities

i.e. $$\cos(\omega t+\phi)+\cos \omega t=2\cos(\phi/2)\cos(\omega t + \phi/2)$$

8. Jun 13, 2014

### alingy1

That is what I had done. I did not think I was going the right way.
So, from this, the temporal and spatial are again separated like the regular derivation.

$$D_R=\sin(kx-\omega t+\phi)\\ D_L=\sin(kx+\omega t)\\ D_R+D_L = \sin(kx-\omega t+\phi) + \sin(kx+\omega t)\\ \qquad = \sin kx \cos(\omega t+\phi) - \cos kx\sin(\omega t + \phi) + \sin kx \cos \omega t + \cos kx\sin \omega t\\ \qquad = \big(\cos(\omega t+\phi)+\cos \omega t \big)\sin kx + \big( \sin \omega t-\sin(\omega t + \phi)\big)\cos kx\qquad$$

$$2\cos(\phi/2)\cos(\omega t + \phi/2)+2\sin(\phi/2)\cos(\omega t + \phi/2)$$
$$=2\cos(\omega t + \phi/2)(\cos(\phi/2)\sin(kx)+\sin(\phi/2)\cos(kx))$$
$$=2\cos(\omega t + \phi/2)(\sin(\phi/2+kx))$$

Again, the famous separation of spatial and temporal in standing waves.
It seems as though all the nodes are shifted to the left by $$(\phi/2)*(\lambda/(2\pi))$$

PS: This is my first LaTex post. I cannot stress how proud I am :) With all seriousness though, is my interpretation correct? And does that mean that we could create standing waves by attaching a string to a wall, shaking one end by making sure the phase constant is phi/2? What would happen to the node at the wall?

9. Jun 14, 2014

### Simon Bridge

That's right - basically, then you added in the original phi you were choosing a particular position to call x=0, as well as when to start your stopwatch. You still get the same standing waves, it's just that the ruler is in a different place.

You don't have to put the x=0 mark on your ruler next to a node - you could put it next to an antinode if you like (that is more natural for pressure waves in a tube for eg) or anywhere in between. You don't have to start the stopwatch (how you get t=0 remember?) when the amplitudes are maximum either.

Since the choice of origin (space and time) is arbitrary, you can choose one that makes the maths simple. I don't know about you but I try to avoid hard maths if I can: a lot of physics (and mathematics) is about trying not to do more maths than you have to.

Notice you also get twice the amplitudes that you had for each initial wave?
You can also explore what happens if the amplitudes of the component waves are not the same.

You can see why everyone insists people learn LaTeX can't you - that is good for a first try.

10. Jun 14, 2014

### alingy1

Yes, perfect. I still have one question I am not really sure of.
When I plot the final function
$$=2\cos(\omega t + \phi/2)(\sin(\phi/2+kx))$$
I see that the nodes are at a separate place.
For the case with the string attached to the wall, does that mean that the wall is no longer a node?

Or am I again just making a reference system mistake?

11. Jun 14, 2014

### Simon Bridge

No - OK you just triggered one of my potted lessons:

What you did above was specify that some arbitrary pair of waves would have a phase difference of $\small{\phi}$ at (x,t)=(0,0) and then explored the consequences. i.e. you started with a pair of waves and asked what the resulting behavior would be...

Thus the equation does not describe the standing waves on a string fixed at x=0. It can be used to describe standing waves on a string fixed at x=<somewhere else> though. This is what I meant when I talked about moving your ruler.

But you could do it the other way. Starting with:
$\qquad y(x,t)=A\cos(\omega t+\alpha)\sin(kx+\beta)$ [1]
... find out what the different constants have to be to give you the behavior that is in front of you.
This approach is more useful.

Examples:

(1) The string is fixed at the wall, and the wall is at x=0, then $y(0,t)=0$
... which tells you that you are looking for equations with $\beta=0$

(2) But maybe you put the wall at x=-L/2 (so the origin is in the middle of the string). [2]
In that case $y(-L/2,t)=0$ so solutions with $\beta = kL/2$ should work.

The end fixed to the wall has to be a node - that's the physics.
Any maths you use to describe it must also have a node at the position of the wall.

Discussion:

I think this is something that gets overlooked:
The position of the origin of the coordinate system is something you choose.

The origin for length is the 0 mark on your ruler.
The origin for time is whenever you happen to start the stopwatch.

You have a lot of control over how you do the maths - you don't have to choose +y to be "up" for example ... it could be "down" or (as in the case for longitudinal waves) "along". Choose what makes sense for what you are trying to find out (preferably what also makes the maths easy.)

The only restriction is this: Nature has to agree with you.
Nature knows more physics that you do - heed what She says.

-----------------------------

[1] This is deliberately different from yours.
You have $\alpha=\beta=\phi/2$ because of the initial conditions you chose.
In general, you should let Nature help you decide which conditions to choose.

The general equation for any old 1D wave looks like this:
$$\frac{\partial^2}{\partial t^2}y(x,t) = c^2\frac{\partial^2}{\partial x^2}y(x,t)$$... where c is the wave speed.

This equation describes waves of any kind in 1D, but it's just maths. To be useful, you need to plug in your knowledge of the physics.

[2] This choice is common if there is a lot of symmetry - like if both ends are fixed.
An example is a particle confined to an infinite square well.

12. Jun 14, 2014

### alingy1

I am going to ask a silly question.
Can I form a standing wave if the wall has a node (as always) but the other end of the string that I am holding is moving?

Thanks a lot. I really appreciate your potted lessons :) Long and detailed to keep the problems away.

13. Jun 15, 2014

### Simon Bridge

Not at all silly.
What you are asking is actually from the next level of your education.

What you have been learning is for free propagating sine waves - so you start out with a sine-wave displacement of the string and it just keeps going without any further input. It says nothing about how you get that distortion in the first place, or what happens if the ideal situation does not happen (which is pretty much all the time).

You will have noticed that if you just pluck a string and let go (sticking to fixed at both ends for now) then distortion starts out as a triangle, quickly settles down to a sine wave - in fact, to the fundamental - then gradually dies down completely.

This is the result of a phenomena called "damping" - the energy from the initial plucking goes to kinetic energy in the string, but also to making sound and heat and moving air around etc. This is like a friction effect - but it is proportional to the frequency so you end up with the lowest frequency wave before everything dies down.

To keep the wave going, you have to add energy. This is where your grabbing the end of the string and shaking it comes in. This is called "driving" or "generating" the wave. The equations are for driven-damped harmonic motion and are quite complicated.

The basic result that is important to your questions is "resonance".
If you drive the string at a frequency close to one of the harmonics for the string, then you get that harmonic appearing. However, the effective length of the string (for determining the harmonic) depends on the amplitude of the driving oscillations - i.e. how much you wave your end up and down. If you have low amplitude oscillations, which is all you need anyway, then you will get resonant standing waves using up almost the whole length of the string.

The best way to see this is to try it out.
School lab equipment sometimes includes a mutivibrator that can be used to set it up carefully but you can just, as you say, attach one end of a string to a wall and shake the other end up and down. The trick is to avoid circular motion, like a jump-rope, in the string.

The way to generate standing waves on a string with only one fixed end, is to wave the "fixed" end.
Small amplitude waving will put a node very close to your hand - that is the "fixed" point.

You pretty much need a stiff string - use a wire - or lay a chain along a desktop and use a back and forth motion (though you'll see a lot of damping that way too).