Propagation Wave Expressions and Wave Velocity

[ashah99]

Homework Statement

How to derive the expressions for the forward and backward propagation waves and to calculate te wave velocity. The problem statement is in the attachment below.

Relevant Equations

For wave velocity: v = omega/k, where w is radial frequency and k is the wavenumber or v = lambda*f, where lambda is the wavelength and f is the frequency

Hello everyone, I would really appreciate some help on the following problem on plane waves and propagation. Not too sure if my attempt at writing the propagation wave expressions are correct, and how to handle the arbitrary function f(u). For the velocity, the wavelength is not specified, so is this how you calculate it? All guidance is appreciated.

The problem statement is given below:
https://physicshelpforum.com/attachments/1629990852598-png.4421/

Attempt at an answer:

https://physicshelpforum.com/attachments/1629990968681-png.4422/

 

[berkeman]

Homework Statement:: How to derive the expressions for the forward and backward propagation waves and to calculate te wave velocity. The problem statement is in the attachment below.
Relevant Equations:: For wave velocity: v = omega/k, where w is radial frequency and k is the wavenumber or v = lambda*f, where lambda is the wavelength and f is the frequency

The problem statement is given below:
https://physicshelpforum.com/attachments/1629990852598-png.4421/

Attempt at an answer:

https://physicshelpforum.com/attachments/1629990968681-png.4422/

Your attachments don't seem to have worked. Please try again, and keep in mind that we prefer that you type your work into the forum window using LaTeX (see the Guide link below the Edit window).

 

[ashah99]

Here is my re-upload with problem and solution attempt.

 

[Steve4Physics]

Here is my re-upload with problem and solution attempt.

Hi.

Your answer to a) is correct though your reasoning is wrong because 'waves' need not be periodic (see below). You have barked up the wrong tree for part b) and, as a result, you haven’t calculated the speed of the backward propagating wave in part c).

See if this helps…

I believe the wording in the question is poor/misleading I will tell you that the given equation is for the sum of 2 waves so the questions should say:
(a) Write an expression for the forward-propagating wave.
(b) Write an expression for the backwards-propgating wave.
(c) Calculate the velocities of these waves [which you will find are erqual].
____________

Don’t use ‘f’ for frequency here. We are told that f(u) is some arbitrary function. You don’t want to get the function and frequency mixed up.
____________

‘Waves’ in the general sense aren’t necessarily sinusoidal or even repetitive (periodic/cyclical). A ‘wave’ is just some pattern moving with a (say constant) speed. (To be a bit more rigorous, a wave is a solution of the wave equation - a partial differential equation which you can research if interested.)

For example, a simple, single, moving pulse (of any shape) may be referred to as a wave – in which case, there would be no wavelength or frequency. An example with one ‘wave’ moving left and another moving right is this: https://scienceworld.wolfram.com/physics/gifs/waves.gif
____________

Q1. Suppose ‘f’ is some (unknown) function. What is the difference between the graphs of y = f(1-x), y = f(2-x) and y = f(3-x)? Nothing complicated, - just think high-school maths.

Q2. Suppose we have y = f(t - x). What would the series of graphs look like for increasing values of t (say t=0, t=1, t=2, t=3, …)?

Q3. Suppose y = f(3t - x). How would that change the series of graphs?

Q4. Suppose y = f(3t + x). How would that change the series of graphs?

Q5. Suppose y = f(3t -x + 100). How would that change the series of graphs?

Q6. Suppose y = f(7(3t – x)). How would that change the series of graphs?
______________

From answering the above questions, you should be able to see why, for example, y =f(vt - x + constant) is a wave moving ‘forwards' (+x direction) with speed v,

Q7. y = f(vt + x) gives a wave in what direction?

(Of course, for propagation in the z-direction, we would use ‘z’ rather than ‘x’)
____________

In the expression $f \Big( 2t + \frac {z}{100} + 12 \Big)$ you are not being told what f(u) is; you are being told what u is. $u(z, t) = 2t + \frac {z}{100} + 12$.

(It doesn't matter what 'f' is. It could be $f(u) = u^2$ or $f(u) = sin(u^3) + \frac {1}{u}$ or anything. It is not relevant.)
____________

When you can answer the above questions, you can revisit your original question!

Hint: With some simple algebra, can you express $u(z,t)$ in a form which contains $vt+z$, where v is some number?

 

[ashah99]

Hi.

Your answer to a) is correct though your reasoning is wrong because 'waves' need not be periodic (see below). You have barked up the wrong tree for part b) and, as a result, you haven’t calculated the speed of the backward propagating wave in part c).

See if this helps…

I believe the wording in the question is poor/misleading I will tell you that the given equation is for the sum of 2 waves so the questions should say:
(a) Write an expression for the forward-propagating wave.
(b) Write an expression for the backwards-propgating wave.
(c) Calculate the velocities of these waves [which you will find are erqual].
____________

Don’t use ‘f’ for frequency here. We are told that f(u) is some arbitrary function. You don’t want to get the function and frequency mixed up.
____________

‘Waves’ in the general sense aren’t necessarily sinusoidal or even repetitive (periodic/cyclical). A ‘wave’ is just some pattern moving with a (say constant) speed. (To be a bit more rigorous, a wave is a solution of the wave equation - a partial differential equation which you can research if interested.)

For example, a simple, single, moving pulse (of any shape) may be referred to as a wave – in which case, there would be no wavelength or frequency. An example with one ‘wave’ moving left and another moving right is this: https://scienceworld.wolfram.com/physics/gifs/waves.gif
____________

Q1. Suppose ‘f’ is some (unknown) function. What is the difference between the graphs of y = f(1-x), y = f(2-x) and y = f(3-x)? Nothing complicated, - just think high-school maths.

Q2. Suppose we have y = f(t - x). What would the series of graphs look like for increasing values of t (say t=0, t=1, t=2, t=3, …)?

Q3. Suppose y = f(3t - x). How would that change the series of graphs?

Q4. Suppose y = f(3t + x). How would that change the series of graphs?

Q5. Suppose y = f(3t -x + 100). How would that change the series of graphs?

Q6. Suppose y = f(7(3t – x)). How would that change the series of graphs?
______________

From answering the above questions, you should be able to see why, for example, y =f(vt - x + constant) is a wave moving ‘forwards' (+x direction) with speed v,

Q7. y = f(vt + x) gives a wave in what direction?

(Of course, for propagation in the z-direction, we would use ‘z’ rather than ‘x’)
____________

In the expression $f \Big( 2t + \frac {z}{100} + 12 \Big)$ you are not being told what f(u) is; you are being told what u is. $u(z, t) = 2t + \frac {z}{100} + 12$.

(It doesn't matter what 'f' is. It could be $f(u) = u^2$ or $f(u) = sin(u^3) + \frac {1}{u}$ or anything. It is not relevant.)
____________

When you can answer the above questions, you can revisit your original question!

Hint: With some simple algebra, can you express $u(z,t)$ in a form which contains $vt+z$, where v is some number?

I'll attempt to answer your exercises. Though this is not my strong area (physics/math)

Q1. Suppose ‘f’ is some (unknown) function. What is the difference between the graphs of y = f(1-x), y = f(2-x) and y = f(3-x)? Nothing complicated, - just think high-school maths.

• So f(-x) would just be a reflection about the y-axis, where all x-values are negated. Then, y = f(1-x), y = f(2-x) and y = f(3-x) are just horizontal shifts of 1, 2, 3 units to the left for f(-x), respectively.

Q2. Suppose we have y = f(t - x). What would the series of graphs look like for increasing values of t (say t=0, t=1, t=2, t=3, …)?

• The graph would shift to the left by t = n units for increasing t?

Q3. Suppose y = f(3t - x). How would that change the series of graphs?

• 3t would be horizontal scaling. y = f( 3[t - x/3] ), so divide every x-coordinate by 3 and move 1/3 horizontal shift to the right

Q4. Suppose y = f(3t + x). How would that change the series of graphs?

• 3t would be horizontal scaling. y = f( 3[t + x/3] ), so divide every x-coordinate by 3 and move 1/3 horizontal shift to the left

Q5. Suppose y = f(3t -x + 100). How would that change the series of graphs?

• Not real sure here when the +100 constant is added

Q6. Suppose y = f(7(3t – x)). How would that change the series of graphs?

• Getting a bit more confused now...

Q7. y = f(vt + x) gives a wave in what direction?

• f(vt + x) represents a backward-propagating wave, or towards the left in the -x direction with velocity v.

u(z, t) can be rewritten as u(z, t) = (1/100)*[200t + z + 1200], but still leaves the 1200.

 

[Steve4Physics]

This is quite difficult stuff, even if you’re fairly good at maths. I think I asked you to do too much at once But you did well! Let’s go through it.
___________

Q1. Suppose ‘f’ is some (unknown) function. What is the difference between the graphs of y = f(1-x), y = f(2-x) and y = f(3-x)? Nothing complicated, - just think high-school maths.

Your answer: So f(-x) would just be a reflection about the y-axis, where all x-values are negated. Then, y = f(1-x), y = f(2-x) and y = f(3-x) are just horizontal shifts of 1, 2, 3 units to the left for f(-x),1 r0espectively.

My answer: I only asked about the difference between the graphs – so don’t worry about the issue of reflection in the y-axis.

Ye, we get horizontal shifts. But the shifts are to the right (‘forwards’). To see this consider a simple example function y = f(x) = 5x.

f(1 - x) means we use 1 - x, rather than x, so we have y = 5(1 - x) = 5 - 5x
f(2 - x) means we use 2 - x, rather than x, so we have y = 5(2 - x) = 10 - 5x
f(3 - x) means we use 3 - x, rather than x, so we have y = 5(3 - x) = 15 - 5x

If you sketch these 3 graphs (y = 5 - 5x, y = 10 – 5x and y = 15 - 5x) you will see we get shifts to the right.
___________

Q2. Suppose we have y = f(t - x). What would the series of graphs look like for increasing values of t (say t=0, t=1, t=2, t=3, …)?

Your answer: The graph would shift to the left by t = n units for increasing t?

My answer: Yes – but to the right. Now you should start seeing how changing time (t in seconds) makes the pattern move.
__________

Q3. Suppose y = f(3t - x). How would that change the series of graphs?

Your answer: 3t would be horizontal scaling. y = f(3[t - x/3]), so divide every x-coordinate by 3 and move 1/3 horizontal shift to the right

My answer: Not quite. In f(3t – x) the ‘3’ is only applied to t, not to x. The effect is that instead of the pattern moving in steps of 1 unit when t increases by 1, the pattern moves 3 units to the right. So if t is time in seconds, the speed of movement is 3 steps per second The ‘3’ is giving us the speed!
___________

Q4. Suppose y = f(3t + x). How would that change the series of graphs?

Your answer: 3t would be horizontal scaling. y = f(3[t + x/3]), so divide every x-coordinate by 3 and move 1/3 horizontal shift to the left

My answer: Not quite. As for my answer to Q3 (but moving to the left).
___________

Q5. Suppose y = f(3t -x + 100). How would that change the series of graphs?

Your answer: Not real sure here when the +100 constant is added

My answer: Same as for Q3, except all graphs are moved 100 units to the right. The speed is still 3 steps per second to the right.
___________

Q6. Suppose y = f(7(3t – x)). How would that change the series of graphs?

Your answer: Getting a bit more confused now…

My answer: That’s a hard question. The 7 changes the shape of the graph (squashes it) but the speed and direction of the (now squashed) graph is still 3 units per second to the right.
___________

Q7. y = f(vt + x) gives a wave in what direction?

Your answer: f(vt + x) represents a backward-propagating wave, or towards the left in the -x direction with velocity v.

My answer: Yes! Spot on. And it doesn’t matter what the function is. The function ‘f’ gives the shape, but having ‘vt + x’ controls the speed and direction the shape moves.
___________

You said: u(z, t) can be rewritten as u(z, t) = (1/100)*[200t + z + 1200], but still leaves the 1200.

My comment: Good. The fact that your expression contains t and z in the form ‘200t + z’ tells you the speed is 200 and the direction is backwards (-z direction). The ‘1/100’ and the ‘1200’ don’t affect the speed and direction (see Q5 and Q6 above).

Incidentally, you might note that cos(1000t – 5z) can be written as cos{5(200t -z)] which gives a simple way to get the speed and direction of that wave. You just look at the '200t - z'.

So I hope you can see that you have now answered the original question. Well done!

Edited: minor changes only.

 

[ashah99]

This is quite difficult stuff, even if you’re fairly good at maths. I think I asked you to do too much at once But you did well! Let’s go through it.
___________

Q1. Suppose ‘f’ is some (unknown) function. What is the difference between the graphs of y = f(1-x), y = f(2-x) and y = f(3-x)? Nothing complicated, - just think high-school maths.

Your answer: So f(-x) would just be a reflection about the y-axis, where all x-values are negated. Then, y = f(1-x), y = f(2-x) and y = f(3-x) are just horizontal shifts of 1, 2, 3 units to the left for f(-x),1 r0espectively.

My answer: I only asked about the difference between the graphs – so don’t worry about the issue of reflection in the y-axis.

Ye, we get horizontal shifts. But the shifts are to the right (‘forwards’). To see this consider a simple example function y = f(x) = 5x.

f(1 - x) means we use 1 - x, rather than x, so we have y = 5(1 - x) = 5 - 5x
f(2 - x) means we use 2 - x, rather than x, so we have y = 5(2 - x) = 10 - 5x
f(3 - x) means we use 3 - x, rather than x, so we have y = 5(3 - x) = 15 - 5x

If you sketch these 3 graphs (y = 5 - 5x, y = 10 – 5x and y = 15 - 5x) you will see we get shifts to the right.
___________

Q2. Suppose we have y = f(t - x). What would the series of graphs look like for increasing values of t (say t=0, t=1, t=2, t=3, …)?

Your answer: The graph would shift to the left by t = n units for increasing t?

My answer: Yes – but to the right. Now you should start seeing how changing time (t in seconds) makes the pattern move.
__________

Q3. Suppose y = f(3t - x). How would that change the series of graphs?

Your answer: 3t would be horizontal scaling. y = f(3[t - x/3]), so divide every x-coordinate by 3 and move 1/3 horizontal shift to the right

My answer: Not quite. In f(3t – x) the ‘3’ is only applied to t, not to x. The effect is that instead of the pattern moving in steps of 1 unit when t increases by 1, the pattern moves 3 units to the right. So if t is time in seconds, the speed of movement is 3 steps per second The ‘3’ is giving us the speed!
___________

Q4. Suppose y = f(3t + x). How would that change the series of graphs?

Your answer: 3t would be horizontal scaling. y = f(3[t + x/3]), so divide every x-coordinate by 3 and move 1/3 horizontal shift to the left

My answer: Not quite. As for my answer to Q3 (but moving to the left).
___________

Q5. Suppose y = f(3t -x + 100). How would that change the series of graphs?

Your answer: Not real sure here when the +100 constant is added

My answer: Same as for Q3, except all graphs are moved 100 units to the right. The speed is still 3 steps per second to the right.
___________

Q6. Suppose y = f(7(3t – x)). How would that change the series of graphs?

Your answer: Getting a bit more confused now…

My answer: That’s a hard question. The 7 changes the shape of the graph (squashes it) but the speed and direction of the (now squashed) graph is still 3 units per second to the right.
___________

Q7. y = f(vt + x) gives a wave in what direction?

Your answer: f(vt + x) represents a backward-propagating wave, or towards the left in the -x direction with velocity v.

My answer: Yes! Spot on. And it doesn’t matter what the function is. The function ‘f’ gives the shape, but having ‘vt + x’ controls the speed and direction the shape moves.
___________

You said: u(z, t) can be rewritten as u(z, t) = (1/100)*[200t + z + 1200], but still leaves the 1200.

My comment: Good. The fact that your expression contains t and z in the form ‘200t + z’ tells you the speed is 200 and the direction is backwards (-z direction). The ‘1/100’ and the ‘1200’ don’t affect the speed and direction (see Q5 and Q6 above).

Incidentally, you might note that cos(1000t – 5z) can be written as cos{5(200t -z)] which gives a simple way to get the speed and direction of that wave. You just look at the '200t - z'.

So I hope you can see that you have now answered the original question. Well done!

Edited: minor changes only.

Thank you for taking the time to explain the concepts. I believe I understand this better now and perhaps I overcomplicated things.

I believe the final answers would be:

(a) expression for forward propagating wave is 60cos(1000t - 5z) = 60cos(5(200t-z)). This is sinusoidal with velocity of 200 m/s moving in the +z direction.

(b) expression for backward propagating wave = -120*pi*f(2t + z/100 +12) = -120*pi*f( (1/00)*(200t + z + 1200)). A wave with arbitrary function with velocity of 200 m/s moving in the +z direction.

(c) velocity for both waves = 200 m/s by inspection when rewritten in the form f(vt + z)

 

[Steve4Physics]

I believe the final answers would be:

(a) expression for forward propagating wave is 60cos(1000t - 5z) = 60cos(5(200t-z)). This is sinusoidal with velocity of 200 m/s moving in the +z direction.

(b) expression for backward propagating wave = -120*pi*f(2t + z/100 +12) = -120*pi*f( (1/00)*(200t + z + 1200)). A wave with arbitrary function with velocity of 200 m/s moving in the +z direction.

(c) velocity for both waves = 200 m/s by inspection when rewritten in the form f(vt + z)

Good - but check the direction you have stated for part b)!

Some final thoughts...

Using $v = \frac {\omega}{k}$ for expressions containing sine or cosine (as you did in your first answer) is perfectly correct and (of course) gives the same answer because $\omega t - kz = k(\frac {\omega}{k}t - z )$.

The question doesn’t state SI units are being used, so you may not be able to say that the speed is in m/s. You could add something such as “(assuming SI units)” if you needed to cover yourself.

It's always worth answering the exact question stated. You are not required to give the speeds for parts a) and b), you are only told to 'write an expression'.

Finally As a general rule, for a wave of fixed shape, propagating at constant speed v in 1D (say the x-direction), we can express the wave as a function of (±vt ±x):
+vt + x gives a wave propagating in the -x direction
+vt - x gives a wave propagating in the +x direction
-vt + x gives a wave propagating in the +x direction
-vt - x gives a wave propagating in the -x direction

Rule:
Same signs (+ + or - -) give negative direction.
Opposite signs (+ - or - +) give positive direction.

 

[ashah99]

Good - but check the direction you have stated for part b)!

Some final thoughts...

Using $v = \frac {\omega}{k}$ for expressions containing sine or cosine (as you did in your first answer) is perfectly correct and (of course) gives the same answer because $\omega t - kz = k(\frac {\omega}{k}t - z )$.

The question doesn’t state SI units are being used, so you may not be able to say that the speed is in m/s. You could add something such as “(assuming SI units)” if you needed to cover yourself.

It's always worth answering the exact question stated. You are not required to give the speeds for parts a) and b), you are only told to 'write an expression'.

Finally As a general rule, for a wave of fixed shape, propagating at constant speed v in 1D (say the x-direction), we can express the wave as a function of (±vt ±x):
+vt + x gives a wave propagating in the -x direction
+vt - x gives a wave propagating in the +x direction
-vt + x gives a wave propagating in the +x direction
-vt - x gives a wave propagating in the -x direction

Rule:
Same signs (+ + or - -) give negative direction.
Opposite signs (+ - or - +) give positive direction.

Yes, my mistake on part b. It would be in the -z direction. That was me being careless by copy pasting. As you mentioned, I agree that it’s best to skip on the extra info and answer the question directly.

final answers would be:

(a) Expression for forward propagating wave is 60cos(1000t - 5z).

(b) Expression for backward propagating wave = -120*pi*f(2t + z/100 +12)

(c) velocity for both waves = 200

thank you for your help and bearing with me throughout the process!

 

[Steve4Physics]

Yes, my mistake on part b. It would be in the -z direction. That was me being careless by copy pasting. As you mentioned, I agree that it’s best to skip on the extra info and answer the question directly.

final answers would be:

(a) Expression for forward propagating wave is 60cos(1000t - 5z).

(b) Expression for backward propagating wave = -120*pi*f(2t + z/100 +12)

(c) velocity for both waves = 200

thank you for your help and bearing with me throughout the process!

You're welcome. I will still add a couple more comments!

In part c) you should include the method(s) of calculation of the speeds. The question says 'calculate'. When asked to calculate something, it is usually requiured for your working to be included in the answer. (If you are simly asked to 'state' something, then no working need be shown.) So you should clearly show how you found the speed of each wave.

I would also add, for example, '(unknown units)' after '200'. Or something on those lines.