Help on reflection of a wave from fixed end without transmission

[mrpurpletoes]

Homework Statement

im having a hard time figuring out how reflection of a wave from fixed end with 0 transmission and 0 energy loss changes the equation of a wave.
i've been told the reflection of the wave y=A sin(wt-Kx) is y=-Asin(wt+Kx) which is obtained by:
- changing the sign of -Kx to +Kx to indicate change in direction (A sin(wt+Kx))
- then multiplying with -1 to indicate inversion of the wave

but when i follow these steps for the wave y=2 sin(4x-8t)
i get 2 sin(4x+8t) when the answer key says it should be -2 sin(4x+8t). Where am i going wrong?

Relevant Equations

y=A sin(wt-Kx)
y=2 sin(4x-8t)

y=2 sin(4x-8t)

y=2 sin(-4x-8t) (opposite direction)

y= 2 sin(4x+8t) (opposite direction and inverted)

 

[A.T.]

y=A sin(wt-Kx)
y=2 sin(4x-8t)

Why did you swap the x and t terms here?

y=2 sin(-4x-8t) (opposite direction)

y= 2 sin(4x+8t) (opposite direction and inverted)

In the last step, you have to negate the entire function, not just some random partial term.

 

[mrpurpletoes]

Why did you swap the x and t terms here?

i took w=-8 and K=-4

In the last step, you have to negate the entire function, not just some random partial term.

could you explain what negating the entire function means

 

[kuruman]

You did not verify that what you wrote down is what you want. What you wrote as "opposite direction"

y=2 sin(-4x-8t) (opposite direction)

is the answer. That's because the sine is an odd function meaning that sin(-u) = -sin(u)
So sin(-4x - 8t) = -sin(4x + 8t) which is what you want.

The convention for x is that it increases to the right and decreases to the left. Time t always increases and frequency w is always positive. If you write y = sin(kx - wt), you see that the wave travels to the right because since time always increases, to keep the phase (kx - wt) constant x must increase. Likewise, If you write y = sin(kx + wt), you see that the wave travels to the left because since time always increases, to keep the phase (kx + wt) constant x must decrease. The same would be true if you wrote the wave as a cosine.

The bottom line is to always write the wave as y = sin(kx ± wt) with the understanding that the top sign is for a wave traveling to the left and the bottom sign for a wave traveling to the right. To flip direction, you flip the sign in front of w. It's a good idea to bring the wave in this standard form before you do any flipping.

For example, if this problem gave you y = sin(-kx - wt) as the initial wave, you would write
y = sin(-kx - wt) = -sin(kx + wt) is the given wave in standard form
y = -sin(kx - wt) is the given wave traveling in the opposite direction
y = +sin(kx - wt) is the inverted given wave traveling in the opposite direction

 

[A.T.]

could you explain what negating the entire function means

Negate means multiply by -1. Look at the solution that you were supposed to get. It has a minus sign in front of the entire expression. This is how you flip a function vertically, which is what want for wave inversion.

 

[Herman Trivilino]

- then multiplying with -1 to indicate inversion of the wave

but when i follow these steps for the wave y=2 sin(4x-8t)

If $y=2 \sin(4x-8t)$ then what is $-y$ equal to?

 

[haruspex]

Homework Statement: im having a hard time figuring out how reflection of a wave from fixed end with 0 transmission and 0 energy loss changes the equation of a wave.
i've been told the reflection of the wave y=A sin(wt-Kx) is y=-Asin(wt+Kx) which is obtained by:
- changing the sign of -Kx to +Kx to indicate change in direction (A sin(wt+Kx))
- then multiplying with -1 to indicate inversion of the wave

but when i follow these steps for the wave y=2 sin(4x-8t)
i get 2 sin(4x+8t) when the answer key says it should be -2 sin(4x+8t). Where am i going wrong?
Relevant Equations: y=A sin(wt-Kx)
y=2 sin(4x-8t)

y=2 sin(4x-8t)

y=2 sin(-4x-8t) (opposite direction)

y= 2 sin(4x+8t) (opposite direction and inverted)

I assume the fixed end is at x=0. That means that the sum of the waves should be always zero there.
Official answer :
2 sin(4x-8t) -2 sin(4x+8t) = -2 sin(8t) -2 sin(8t) = -4 sin(8t)
Your answer:
2 sin(4x-8t) +2 sin(4x+8t) = -2 sin(8t) +2 sin(8t) = 0.

You win.

 

[A.T.]

I assume the fixed end is at x=0. That means that the sum of the waves should be always zero there.
Official answer :
2 sin(4x-8t) -2 sin(4x+8t) = -2 sin(8t) -2 sin(8t) = -4 sin(8t)
Your answer:
2 sin(4x-8t) +2 sin(4x+8t) = -2 sin(8t) +2 sin(8t) = 0.

You win.

The official answer formula for the reflected wave is:
y=-Asin(wt+Kx)
which using these values:

i took w=-8 and K=-4

and A = 2 becomes:
y =-2 sin(-8t-4x) = 2 sin(8t+4x)
Which is the same as the OPs answer.

...the answer key says it should be -2 sin(4x+8t).

ETA: If that's the explicit numerical answer key, then it's indeed wrong.

 

[haruspex]

The official answer for the refected wave is:
y=-Asin(wt+Kx)

That’s not what it says in post #1. It says that is the formula that he had been taught to use, but the answer provided to the question was -2 sin(4x+8t).

 

[A.T.]

That’s not what it says in post #1. It says that is the formula that he had been taught to use, but the answer provided to the question was -2 sin(4x+8t).

OK, then I wrongly assumed he got -2 sin(4x+8t) by plugging his parameters into the formula.

 

[DaveE]

That means that the sum of the waves should be always zero there.

Careful. This depends on the nature of the problem; the boundary conditions. Reflection coefficients can be positive or negative. I guess he was told that this one sums to zero, but that's not always true.

 

[Herman Trivilino]

I guess he was told that this one sums to zero

He never showed us the statement of the problem he was asked to solve.

 

[DaveE]

He never showed us the statement of the problem he was asked to solve.

No, but he was told the reflected wave travelled backwards and was inverted. So, $\Gamma = -1$. My concern would be that he thinks this is always true. It's not.

 

[haruspex]

Careful. This depends on the nature of the problem; the boundary conditions. Reflection coefficients can be positive or negative. I guess he was told that this one sums to zero, but that's not always true.

"reflection of a wave from fixed end"

 

[haruspex]

OK, then I wrongly assumed he got -2 sin(4x+8t) by plugging his parameters into the formula.

I see no need to be making any assumptions. Post #1 says he got +2 sin(4x+8t) by plugging his parameters into the formula.

 

[A.T.]

Post #1 says he got +2 sin(4x+8t) by plugging his parameters into the formula.

Not quite. Post #1 says he got +2 sin(4x+8t) by following the steps that lead to the formula with his numbers. So it sounded to me like he was checking the consistency between the steps and the formula, using his numbers.

 

[haruspex]

Not quite. Post #1 says he got +2 sin(4x+8t) by following the steps that lead to the formula with his numbers. So it sounded to me like he was checking the consistency between the steps and the formula, using his numbers.

So what would "answer key" refer to in that reading?

 

[A.T.]

So what would "answer key" refer to in that reading?

I already explained my misunderstanding in post #10.

 

[kuruman]

I assume the fixed end is at x=0.

The problem does not specify where the string is tied. Both solutions can be correct. Using the identity $$\sin a \pm\sin b=2\sin\frac{a\pm b}{2}\cos\frac{a\mp b}{2},$$ we find $$y_{\text{OP}}=4\sin(4x)\cos(8t)~;~~y_{\text{key}}=-4\cos(4x)\sin(8t).$$ Both expressions give standing waves of the same wavelength except that their nodes occur at different places on the x-axis. The statement of the problem is ambiguous.

 

[Herman Trivilino]

I see no need to be making any assumptions.

You are assuming that the original problem asks for the equation of the wave. But I was assuming it was asking for the equation of the reflected portion of the wave.

We can (probably) safely assume that your assumption is correct based on the answers given by the OP and the answer key.

But I still maintain that we never saw the original problem statement. Which leads to confusion and at the very least temporarily and unnecessarily misdirecting the OP.

 

[DaveE]

The problem does not specify where the string is tied.

It doesn't even specify it's a string. Hence my previous comments.
This is 99% simple trigonometry and 1% physics.
We've done the math part, time to move on I think.

 

[Herman Trivilino]

The problem does not specify where the string is tied.

There is no statement of the problem. How could there be a specification of anything?

 

[haruspex]

The problem does not specify where the string is tied. Both solutions can be correct. Using the identity $$\sin a \pm\sin b=2\sin\frac{a\pm b}{2}\cos\frac{a\mp b}{2},$$ we find $$y_{\text{OP}}=4\sin(4x)\cos(8t)~;~~y_{\text{key}}=-4\cos(4x)\sin(8t).$$ Both expressions give standing waves of the same wavelength except that their nodes occur at different places on the x-axis. The statement of the problem is ambiguous.

As I wrote in post #7:

I assume the fixed end is at x=0.

In post #15, I should have been more specific: I meant the sort of assumption A.T. was making about where the conflicting answers +2 sin(4x+8t) and -2 sin(4x+8t) came from.
But you make an interesting point, that if the question is incompletely quoted in post #1 then perhaps the fixed point was given as somewhere else. However, that does strike me as unlikely; "at x=0” is more likely not to have been mentioned (whether by the problem setter or the OP) than any other specification.

You are assuming that the original problem asks for the equation of the wave. But I was assuming it was asking for the equation of the reflected portion of the wave.

Eh? No, everyone agrees it is asking for the equation of the reflected wave, and that is what the OP correctly found. The total wave would be standing, trig function of position times trig function of time.

 

[Herman Trivilino]

Eh? No, everyone agrees it is asking for the equation of the reflected wave, and that is what the OP correctly found. The total wave would be standing, trig function of position times trig function of time.

Okay, I made a mistake thinking that.

But if $y = 2 \sin(4x-8t)$ is a wave on a string traveling to the right and it reflects off a barrier fixed at $y=0$ then the reflected wave must be $y = -2 \sin(4x+8t)$.

It seems to me that in your Post #7 there has got to be an error.

I assume the fixed end is at x=0. That means that the sum of the waves should be always zero there.
Official answer :
2 sin(4x-8t) -2 sin(4x+8t) = -2 sin(8t) -2 sin(8t) = -4 sin(8t)
Your answer:

2 sin(4x-8t) +2 sin(4x+8t) = -2 sin(8t) +2 sin(8t) = 0.

You win.

It seems to me that you're saying that $y = 2 \sin(4x-8t)$ is a wave on a string traveling to the right and it reflects off a barrier fixed at $y=0$ then the reflected wave is not $y = -2 \sin(4x+8t)$.

Firstly, I don't see how a wave travelling to the right could reflect off a point where $x=0$.

Secondly, perhaps the refection is not at a point where $y=0$?

Perhaps there's an error in the OP. That's why I'd like to see a statement of the problem that's being solved.

 

[haruspex]

But if $y = 2 \sin(4x-8t)$ is a wave on a string traveling to the right and it reflects off a barrier fixed at $y=0$

No, you seem to have x and y confused. x is a position along the string and y is its local displacement at position x and time t.
The barrier is at x=0 (we have to assume) and its nature is a fixed point, i.e. y=0 there for all t. So if we substitute x=0 in the equations then the total wave must be zero for all t.
That is not true if the reflected wave is $y = -2 \sin(4x+8t)$ since both waves reduce to $y = -2 \sin(8t)$, but it is true for $y = +2 \sin(4x+8t)$. So $y = -2 \sin(4x+8t)$ cannot be a solution, whereas $y = +2 \sin(4x+8t)$ could be.

 

[Herman Trivilino]

No, you seem to have x and y confused. x is a position along the string and y is its local displacement at position x and time t.

No. I was thinking that the value of $y$ need not be zero at the point where the string reflects, but I now see that's nonsense.

What had, like a worm, infected my brain was this mistaken claim by the OP:

i've been told the reflection of the wave y=A sin(wt-Kx) is [...] obtained by [...] multiplying with -1 to indicate inversion of the wave

The two waves travelling in opposite directions, described by $y=A \sin (kx-\omega t)$ and $y=A \sin (kx+\omega t)$, will produce a standing wave. One would not write $y=-A \sin (kx+\omega t)$ for the equation of the reflected wave.

If a pulse of height $y=2$ reflects off a fixed end, the returning pulse does indeed have a height $y=-2$. But to see that one has to wait for a time $t$ equal to one quarter of the wave's period. Again, it doesn't imply that one would write $y=-A \sin (kx+\omega t)$ for the equation of the reflected way.

My mistake.

 

[haruspex]

What had, like a worm, infected my brain was this mistaken claim by the OP:

You omitted this step:
"changing the sign of -Kx to +Kx to indicate change in direction (A sin(wt+Kx))"

The two waves travelling in opposite directions, described by $y=A \sin (kx-\omega t)$ and $y=A \sin (kx+\omega t)$, will produce a standing wave. One would not write $y=-A \sin (kx+\omega t)$ for the equation of the reflected wave.

It's a bit confusing because the standard procedure the OP quotes takes the original wave as $y=A \sin (\omega t-kx)$, not $y=A \sin (kx-\omega t)$. Applying the two steps does correctly produce $y=-A \sin (kx+\omega t)$ for the equation of the reflected wave. But the actual problem to be solved has the original wave as y=2 sin(4x-8t), so to apply the procedure as quoted we have to substitute k=-4, $\omega=-8$, leading to a reflected wave y= 2 sin(4x+8t).

If a pulse of height $y=2$ reflects off a fixed end, the returning pulse does indeed have a height $y=-2$. But to see that one has to wait for a time $t$ equal to one quarter of the wave's period.

I don't see that. When the "in principle" wave height of 2 reaches the fixed point then, at that instant, the reflected wave must have height -2 at that point.

 

[Herman Trivilino]

You omitted this step:
"changing the sign of -Kx to +Kx to indicate change in direction (A sin(wt+Kx))"

Yes. At the time it I considered it a distraction from my point about changing the sign of $y$ being a fundamental error.

It's a bit confusing because the standard procedure the OP quotes takes the original wave as $y=A \sin (\omega t-kx)$, not $y=A \sin (kx-\omega t)$.

Agreed. When the OP stated this ...

i've been told the reflection of the wave y=A sin(wt-Kx) is y=-Asin(wt+Kx) which is obtained by:
- changing the sign of -Kx to +Kx to indicate change in direction (A sin(wt+Kx))
- then multiplying with -1 to indicate inversion of the wave

... as I believe has already been pointed out, it is not consistent with what he did.

but when i follow these steps for the wave y=2 sin(4x-8t)
i get 2 sin(4x+8t)

Which side tracked the first four responses. It wasn't until your (fifth) response that it got straightened out.

And notice that after that the OP disappeared. Once he knew he'd made the right answer he lost interest.

For a physics instructor this is a major problem. Students are understandably interested in answer-making. After all, that's how their success in the course is measured. But instructors want students to instead be interested in sense-making, and that makes for a challenge. In this case the OP seems to have no interest in making sense of the situation, only in making the right answer.

This is why, in my opinion, if the statement of the problem is not given in the template, it's the first thing we should ask for before we start guiding the student towards an understanding of the physics involved in solving the problem.

If a pulse of height $y=2$ reflects off a fixed end, the returning pulse does indeed have a height $y=-2$. But to see that one has to wait for a time $t$ equal to one quarter of the wave's period.

I don't see that. When the "in principle" wave height of 2 reaches the fixed point then, at that instant, the reflected wave must have height -2 at that point.

As an example, think about stretching a rope or the like, with one end attached to a hook in the wall and you holding the other end in your hand. You stretch the rope taut and "in principle" it has a height $y=0$ all along its length. Then you send a pulse of height $y=2$ towards the wall. When that pulse hits the wall, the value of $y$ at the wall is still zero. After all, that's what it means to have a fixed end. After reflection, the pulse has a height of $y=-2$ but that is not at the location of the wall. It starts to form there, but it's not fully formed until after a time $t$ equal to one-fourth of the period, at a value of $x$ located one-fourth of a wavelength away from the wall. Of course, to even speak of a period and wavelength one must assume you will continue to send repeated pulses of the same height, equally spaced in time, so that the standing wave can be created.

Alternatively, you can attach the end of the rope, not to a hook, but to a vertical section of pipe attached to the wall, with a loop in the end of the rope so that it's free to slide up and down the pipe. Again, in principle, you've created not a fixed end but a free end. Now when the pulse hits the end of the rope the reflected pulse will not be inverted, and will be fully formed by the time it begins to return towards your hand.

 

[kuruman]

I will try to summarize what has been said with a plot that, I hope, clarifies fuzzy concepts. We have established that the equation $$y(x,t)=A\sin(kx-\omega t)+A\sin(kx+\omega t)$$ (a) consists of two traveling waves, one to the right and one to the left and (b) it may describe a reflection at $x=0$ where $y(0,t)$ vanishes for all $t$.

What has not been clarified is the context of "reflection". In order to have a reflection, there must be an incident wave and a reflected wave. Which is which depends on one's choice. So let us choose the incident wave coming from the left in the "allowed" region $x<0$. It bounces off at $x=0$ and returns where it came from. The region $x>0$ is "forbidden". The figure below shows a snapshot of (a) a traveling wave to the right (red), (b) a traveling wave to the left (blue) and (c) their sum (black) which is a standing wave. These waves are drawn extended into the shaded forbidden region to assist the reader's visualization. We may imagine them as invisible "ghosts."

We now consider an arbitrary constant-phase point $\text{A}_1$ on the incident red wave at $x=-x_0$ and at $t=t_0.$ The displacement is $y(\text{A}_1)=\sin\left[k(-x_0)-\omega t_0\right]=-\sin(kx_0+\omega t_0).$ This constant-phase point travels to the right with velocity $v=\omega/k$. Similarly, there is a conjugate constant-phase point $\text{A}_2$ on the blue wave in the forbidden region at $x=+x_0$ traveling to the left with the same velocity $v$. The displacement is $y(\text{A}_2)=\sin(kx_0+\omega t_0).$ Clearly, $y(\text{A}_1)+y(\text{A}_2)=0$ for all conjugate pairs $\text{A}_1$ and $\text{A}_2$ at all times.

We can now imagine point $\text{A}_1$ approaching $x=0$ at constant speed. Its inverted ghost reflection $\text{A}_2$ is also approaching $x=0$ from the opposite direction in the forbidden region. The two reach $x=0$ simultaneously at which time incident constant-phase $\text{A}_1$ enters the forbidden region as a ghost whilst $\text{A}_2$ begins its real existence as a reflected and inverted (relative to $\text{A}_1$) constant-phase traveling to the left with the blue wave.

It is worth noting that if an arbitrary waveform is assembled as a linear combination of sinusoidals, then it will also undergo inversion upon reflection as each of its components does so individually.

On edit:
Added shading to the forbidden region to a wave incident from the left. For incidence from the right, one needs just to swap "forbidden" and allowed.

 

[haruspex]

it is not consistent with what he did.

Yes it is.
The given wave is y=2 sin(4x-8t).
The quoted procedure is

"the reflection of the wave y=A sin(wt-Kx) is y=-Asin(wt+Kx) … is obtained by:​

- changing the sign of -Kx to +Kx to indicate change in direction (A sin(wt+Kx))​

- then multiplying with -1 to indicate inversion of the wave"​

which is also correct.
Applying that, reflection is:

y=-2 sin(-4x-8t)​

which the OP simplified to

y=2 sin(4x+8t)​

which is the correct answer.

When that pulse hits the wall, the value of y at the wall is still zero. After all, that's what it means to have a fixed end.

Let’s clarify the discussion by using three different variables for the displacement: the original wave is y, the reflected wave is y' and the sum is y".
y=2 sin(4x-8t) everywhere.
At the fixed point, y=2 sin(4x-8t)=2 sin(-8t).
The reflection is y'=2 sin(4x+8t) everywhere before the fixed point.
(The transmitted wave is -2 sin(4x-8t) everywhere beyond the fixed point. Note this is the same as y' but with x negated, so cancels y everywhere.)
The physically observed wave is the sum of the two, y"=y+y'. At x=0, y+y'=2 sin(-8t)+2 sin(+8t) = 0, as required.

And notice that after that the OP disappeared. Once he knew he'd made the right answer he lost interest.

I suspect he also was discouraged by nearly all the responders telling him he'd made a mistake when he hadn't.

 

[mrpurpletoes]

And notice that after that the OP disappeared. Once he knew he'd made the right answer he lost interest.

actually i disappeared cuz i asked the question a day before my exam and didnt have the time to read all the replies.

thanks for the help everyone