# Equation of reflected wave (fixed end/free end)

I have an equation of a wave as $y = 2 \sin\left( \dfrac{\pi}{6}x - \dfrac{\pi}{4}t \right)$. I want to find the equation of the wave which is formed when it gets reflected from (i) a fixed end or (ii) a free end

So, first of all, the wave will now travel in negative direction, so I have to put a negative sign, either in the $\omega t$ term or the $kx$ term. Then for a fixed end I have to introduce a phase difference of $\pi$. So I'll put the negative sign in front of the equation.

Where do I put the negative sign to change the direction of the wave's velocity? In front of $kx$ or $\omega t$? Also, does this sign depend on the distance after which I have fixed the end or left the end free?

Following is a simple observation:

Red => $y = \sin( x - t )$ [moving right]

Blue => $y = \sin( x + t )$ [moving left]

Green => $y = -\sin( x + t )$ [moving left]

Red + Green => antinode at $x=0$ [standing 1]

Red + Blue => antinode at $x=4.5$ [standing 2]

So, I have free end and fixed end at different x for these two standing waves. Then, how can I decide which one is correct?

• Note that $-\sin(x+t)=\sin(x+t+\pi)$ meaning you get a phase shift of $\pi$ additionally to the reversal of direction. Hence it would be the reflected wave for the fixed end. Sep 8 '18 at 12:17

You have to put it in front $\omega t$ since $$\sin(kx-\omega t)$$ will give you a wave travelling to the right (with respect to the time $t$). Then $$\sin(kx+\omega t)$$ is a wave travelling to the left. You can see this by fixing the location $x$ and vary the time $t$.

The sign does not depend on the distance where you fix the end. The wave will travel to the end and then will be reflected.

• If I put infront of x: sin(-x-wt) -> -sin(x+wt) <- Travels in -ve x direction. Sep 6 '18 at 18:46
• @YashMittal But you introduced a phase shift. Sep 6 '18 at 18:49
• @YashMittal You essentially calculated the reflection at a fixed end $\sin(x+\omega t+\pi)=-\sin(x+\omega t)$. Sep 6 '18 at 19:01
• Here only I have confusion. I used DESMOS app to take a look at the moving graphs. If I chose a point of end at /lambda / 4 and /lambda , then one case will have phase difference in one equation and other case will have no phase difference in same equation. Sep 6 '18 at 19:02
• @YashMittal Maybe you can attach photos of this to your question. Sep 6 '18 at 19:30

$y(x,t)=\sin(\omega\,t -k\, x)$ and $y(x,t)=\sin(-\omega\,t +k\, x)$ are examples of the wave equation for a wave travelling in the positive x-direction.

$y(x,t)=\sin(\omega\,t +k\, x)$ and $y(x,t)=\sin(-\omega\,t -k\, x)$ are examples of the wave equation for a wave travelling in the negative x-direction.

You can change the phase by adding or subtracting a phase angle $\phi$.
For example $y(x,t)=\sin(\omega\,t -k\, x+\phi)$ or $y(x,t)=\sin(\omega\,t -k\, x-\phi)$

• What should be the answer in my case? Sep 7 '18 at 6:53
• @YashMittal Note that Physics is not a homework help site. We generally discourage and/or remove complete answers to homework-like questions. This answer in its present form gives you the information that you need to figure out a solution to your problem.
– rob
Sep 8 '18 at 18:42

at a fixed (hard) boundary, the displacement remains zero and the reflected wave changes its polarity (undergoes a $180^\circ{}$ phase change) and direction opposite (replace $\omega t$ by $-\omega t$)

so,in this case reflected wave will be $y=-2\ sin \left(\dfrac{\pi}{6}x+\dfrac{\pi}{4} t\right)$ because $sin(x+\pi)=-sinx$

at a free (soft) boundary end, reflected wave will not be inverted and rest will be same as above

so, in this case reflected wave will be $y=2 \ sin \left(\dfrac{\pi}{6}x+\dfrac{\pi}{4} t\right)$

above, i've assumed boundary as $x=0$ if boundary was $x=x_{0}$ replace everywhere $x$ by $x-x_{0}$

• Why we can't put '-' sign infront of kx term? This will also give wave travelling in -ve direction. Sep 8 '18 at 14:45
• In the attached figures, sin(kx+wt) and -sin(kx+wt) combined with sin(kx-wt), both are giving standing waves. But, sin(kx+wt) + sin(kx-wt) is giving antinode (free end) at x=0 while other combination is giving antinode at x=4.5. This means, both combination will give free end reflection. Sep 8 '18 at 14:52

If you don't want to be confused by all these signs, just write down the general solution of your problem: $$y(x,t)=A\sin(kx-\omega t ) + B\sin(kx+\omega t)$$ In your case $$A=2$$. Stating that a reflection occurs means that $$B$$ is not null. If reflection arises at $$x=l$$, just solve for B in the proper equation $$y(x=l,t)=0$$ or $$\frac{\partial y}{\partial x} (x=l,t) =0$$ depending on your boundary conditions.