How Does Phase Change Affect the Equation of a Stationary Wave?

[songoku]

TL;DR Summary

This is note I got from my teacher.

Let say a horizontal string of length L is tied on one end to a fixed point and the other end of the string is vibrated so that a progressive wave is formed on the string, travels to the fixed point, being reflected back then superposes with incoming wave.

The equation of incoming wave is $y_1=A \sin(\omega t - kx)$ and the equation of the reflected wave will be $y_2=-A \sin (\omega t+kx)$ since the wave travels in opposite direction and undergoes phase change of $\pi$

The equation of stationary wave will be:
$$y_s=y_1+y_2$$
$$=A \sin(\omega t - kx) -A \sin (\omega t+kx)$$
$$=-2A \cos (\omega t) \sin (kx)$$

My questions
1) Is the value of $t$ in $y_1$ and $y_2$ the same? I think the value will be different because $t$ in $y_1$ is measured from when the string is vibrated and $t$ in $y_2$ is measured when the incoming wave is reflected

2) Same question as question (1) but about $x$. I also think $x$ in $y_1$ and $y_2$ are different because in $y_1$, $x$ is horizontal distance of a point on the wave measured from the source of vibration and in $y_2$ the distance is measured from fixed end

3) The equation $y_1$ is for progressive wave traveling to the right but I found from other source (pdf file I downloaded from internet), equation for wave traveling to the right is $y=A \sin (kx - \omega t)$, so which one is correct? I tried changing the equation a little bit to $y=-A \sin (\omega t - kx)$ but then I got different answer for the question: at the start, in which direction will the wave move, positive or negative vertical direction?

Thanks

 

[anuttarasammyak]

1) You say "when the string is vibrated". I assume the free end should keep vibrating to make standing wave. Two t should be same so that $y_s=0$ at fixed point x=0 anytime.

2) Two different waves y_1 and y_2 superposed at x measured from the fixed points where x=0 is $y_s$. x is common for the two waves to mention where we observe the waves.

3) Right or left is ambiguous because you do not explain positive x direction starting from the fixed point Origin go right or go left. y_1 is going positive x direction. y_2 is going negative x direction, reverse, in your formula.

 

[songoku]

1) You say "when the string is vibrated". I assume the free end should keep vibrating to make standing wave. Two t should be same so that $y_s=0$ at fixed point x=0 anytime.

2) Two different waves y_1 and y_2 superposed at x measured from the fixed points where x=0 is $y_s$. x is common for the two waves to mention where we observe the waves.

1) Yes I also assume the free end should keep vibrating. So $t$ is time measured from the instant the free end is vibrated?

The reason I think $t$ is different for $y_1$ and $y_2$ is because the two waves do not start at the same time. Wave $y_1$ needs to travel along the string for certain amount of time to reach the fixed end and when reflected, $y_2$ is produced so I think $t$ in $y_2$ is when $y_2$ is already produced while $t$ in $y_1$ is measured starting when the string is vibrated at free end

2) So $x$ in $y_s$ is measured from fixed end but $x$ in $y_1$ is measured from source of vibration (free end of the string) because I think $x$ in equation of progressive wave is horizontal distance measured from source of vibration?

3) Right or left is ambiguous because you do not explain positive x direction starting from the fixed point Origin go right or go left. y_1 is going positive x direction. y_2 is going negative x direction, reverse.

Let say the right end of horizontal string is tied to fixed end and I hold the left end then move it up and down starting from equilibrium point to create sinusoidal wave moving to the right and I take right as positive direction. If I want to write down the equation of the progressive wave, what would it be:
$$y_1 = A \sin (\omega t - kx)$$

$$\text{or}$$

$$y_2 = A \sin (kx - \omega t)$$

Thanks

 

[anuttarasammyak]

1)2) You seem to think each $y_1,y_2$ has its own t and x. If so you should distinguish it by say $t_1,t_2,x_1,x_2$ and show us how to translate it to common t and x. But here is the way that $t,x$ does not belong to waves. Waves show how they behave with common $t,x$. For an example x=0 means for both $y_1$ and $y_2$ that here is the fixed point.

3) $y_2$. It goes x-positive direction. but I prefer with minus sign on amplitute
$y_2=-A\sin(kx-\omega t)$ as counter attack against
$y_1=A\sin(-kx-\omega t)$ which goes x-negative direction toward the origin.

 

[Dale]

Is the value of t in y1 and y2 the same?

Unless your teacher is very mean to their students, it is the same. If it were different then it would need to be labeled different, e.g. $t_1$ and $t_2$. By labeling them the same the teacher implies $t=t$.

 

[songoku]

Unless your teacher is very mean to their students, it is the same. If it were different then it would need to be labeled different, e.g. $t_1$ and $t_2$. By labeling them the same the teacher implies $t=t$.

1)2) You seem to think each $y_1,y_2$ has its own t and x.

Yes I have impression the value of $t$ for $y_1$ and $y_2$ are different. It is more about my own interpretation after reading the note from my teacher and I am really confused

I will try to use random numerical values to clear my confusion.

5 meters of horizontal string is labelled P on the left end of it and Q on right end of it. Point Q is tied to a fixed end and point P is vibrated so a progressive wave travels from point P to Q and being reflected back at point Q. The progressive wave has amplitude of 2 cm, frequency of 5 Hz and wavelength of 10 cm so the equation of progressive wave is $y_1=2 \sin (10\pi t - 20\pi x)$ where $t$ is time measured from the start of vibration at point P and $x$ is horizontal distance of point on the wave measured from P

Wave $y_1$ will need 10 seconds to travel from A to B so in this time interval (0 to 10 s) the reflected wave still does not exist. At t = 10 s, $y_1$ is reflected and now there are two waves, $y_1$ and $y_2$ where the equation of $y_2$ will be: $y_2=-2 \sin (10\pi t + 20\pi x)$ and the stationary wave formed by $y_1$ and $y_2$ will be: $y_s=y_1+y_2=-4 \cos (10\pi t) \sin(20\pi x)$

Since $y_2$ exists when $y_1$ is reflected at B, my interpretation (after discussing with friends) is that the value of $t$ in $y_1$ and $y_2$ is not the same, like if I substitute $t=3~s$ into $y_1$, I can not put $t=3~s$ to $y_2$ because at that time $y_2$ does not exist. So let say I want to find the displacement of point R at horizontal distance 1 m from point A and after A has been vibrated for $t=3~s$, I think I can not use equation $y_s$ because stationary wave has not been formed at $t=3~s$ and I have to use equation of $y_1$ by putting $t=3~s$ and $x=1~m$

I apply the same reasoning if I want to find the displacement of point S at 3 m from A after A has been vibrated for 15 s. To reach S from B, $y_2$ needs 6 s so stationary wave will be formed at S when t = 16 s and to find the displacement at S, I can't use $y_s$ and have to use $y_1$ by putting t = 15 s and x = 3 m

Is my interpretation correct?
(a) If yes, does it mean the value of $t$ I can use for $y_2$ and $y_s$ is $t \geq 10$?

(b) If not, what is the mistake in my reasoning?

Thanks

 

[Dale]

If not, what is the mistake in my reasoning?

First, is your teacher a devious or a mean person? If not, then the mistake is to pretend that the teacher is mean when they are not.

Second, usually the standing wave equation is describing a "steady state" scenario well after the initial transients have died down. The details about how the standing wave is originally formed are not relevant and any transient discrepancies from the initial setup have dissipated. We are not starting t=0 at the initiation of the first pulse, but that first pulse was some arbitrarily large negative number that we don't even care about and can just assume is $-\infty$. In fact, that is already what is implied in the beginning with the definition of $y_1=A \sin(\omega t - kx)$. Notice that it is not defined as:
$$y_1=\begin{cases}
A \sin(\omega t - kx) & x>0 \\
0 & \text{Else}
\end{cases}$$
The assumption is already that it has been going on forever, not starting at t=0 which is something that you are just adding on your own.

 

[songoku]

First, is your teacher a devious or a mean person? If not, then the mistake is to pretend that the teacher is mean when they are not.

No, he is not

Second, usually the standing wave equation is describing a "steady state" scenario well after the initial transients have died down. The details about how the standing wave is originally formed are not relevant and any transient discrepancies from the initial setup have dissipated. We are not starting t=0 at the initiation of the first pulse, but that first pulse was some arbitrarily large negative number that we don't even care about and can just assume is $-\infty$.

Ok I get this. So usually in standing wave equation, $t=0~s$ is when the standing wave is already formed and $x$ is horizontal distance measured from a reference point (which is A in my case, not measured from B)

I am curious what will happen to the reflected wave when it reaches A (in the case I wrote). Will it die down when it hits my hand at A (and maybe the energy is dissipated as heat in my hand) or will it be reflected again by my hand and forms another wave $y_3$ which has exactly same equation as $y_1$ or maybe something else?

Thanks

 

[vanhees71]

Let's just do the calculation. The wave equation reads
$$\frac{1}{c}^2 \partial_t^2 u(t,x)-\partial_x^2 u(t,x)=0.$$
The initial condition is
$$u(t=0,x)=0$$
and the boundary conditions
$$u(t,x=0)=0, \quad u(t,x=L)=A \sin(\omega t).$$
The general solution is
$$u(t,x)=u_{L}(x-ct)+u_{R}(x+ct)$$
with arbitrary function $u_L$ and $u_R$. The first boundary condition tells you
$$u(t,0)= \; \Rightarrow u_R=-u_L \; \Rightarrow \; u(t,x)=u_L(x-ct)-u_L(x+ct).$$
This also fulfills the initial condition. From the 2nd boundary condition we get
$$u(t,L)=u_L(L-ct)-u_L(L+ct)=A \sin(\omega t). \qquad (*)$$
Thus we make the ansatz of a standing wave
$$u(t,x)=\sin(\omega t) f(x).$$
From the wave equation we get
$$\frac{1}{c^2} \partial_t^2 u(t,x)-\partial_x^2 u(t,x)=\left [-\frac{\omega^2}{c^2} f(x)-f''(x) \right] \sin(\omega t)=0$$
From this we have
$$f''(x)=-k^2 f(x), \quad k=\frac{\omega}{c}.$$
The general solution is
$$f(x)=C_1 \cos(k x) + C_2 \sin(k x).$$
From $u(t,0)=0$ we find $C_1=0$ and from (*)
$$C_2 \sin(k L)=A \; \Rightarrow \; C_2=\frac{A}{\sin(k L)}.$$
So our final solution is
$$u(t,x)=\frac{A}{\sin(k L)} \sin(\omega t) \sin(k x).$$
That's compatible with our general ansatz, because
$$\sin(\omega t) \sin(k x)=\frac{1}{2} [\cos(k x-\omega t) -\cos(k x+\omega t)]$$
and $\omega = c k$. Thus we have
$$u_L(x)=\frac{A}{2} \cos(k x).$$

 

[aclaret]

The first boundary condition tells you
$$u(t,0)= \; \Rightarrow u_R=-u_L \; \Rightarrow \; u(t,x)=u_L(x-ct)-u_L(x+ct).$$

please may i seek clarification on one thing :) ... if $u(t,0) = u_L(-ct) + u_R(ct) \overset{!}{=} 0$, then how conclude that $u_R = -u_L$? clear that $u_L(0) = - u_R(0)$, but how you can conclude that $u_L(\gamma) = - u_R(\gamma)$ for all $\gamma \in R$?

 

[vanhees71]

Argh: Of course I should have written my ansatz as
$$u(t,x)=u_L(ct-x)+u_R(c t+x).$$

 

[anuttarasammyak]

I am curious what will happen to the reflected wave when it reaches A (in the case I wrote). Will it die down when it hits my hand at A (and maybe the energy is dissipated as heat in my hand) or will it be reflected again by my hand and forms another wave y3 which has exactly same equation as y1 or maybe something else?

Animation in https://en.wikipedia.org/wiki/Standing_wave shows at the antinodes of synthesized waves where one vibrates the string, the reflected waves enhance his effort. So with less effort of waving he would get enhanced amplitude with the help of reflected waves.

 

[songoku]

Thank you very much for the help and explanation anuttarasamyak, Dale, vanhees71, aclaret