4
$\begingroup$

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?

Improve this question
$\endgroup$
1
  • $\begingroup$ 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. $\endgroup$
    – EuklidAlexandria
    Commented Sep 8, 2018 at 12:17

4 Answers 4

Reset to default
2
$\begingroup$

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.

Improve this answer
$\endgroup$
5
  • $\begingroup$ If I put infront of x: sin(-x-wt) -> -sin(x+wt) <- Travels in -ve x direction. $\endgroup$
    – Yash Mittal
    Commented Sep 6, 2018 at 18:46
  • $\begingroup$ @YashMittal But you introduced a phase shift. $\endgroup$
    – EuklidAlexandria
    Commented Sep 6, 2018 at 18:49
  • $\begingroup$ @YashMittal You essentially calculated the reflection at a fixed end $\sin(x+\omega t+\pi)=-\sin(x+\omega t)$. $\endgroup$
    – EuklidAlexandria
    Commented Sep 6, 2018 at 19:01
  • $\begingroup$ 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. $\endgroup$
    – Yash Mittal
    Commented Sep 6, 2018 at 19:02
  • $\begingroup$ @YashMittal Maybe you can attach photos of this to your question. $\endgroup$
    – EuklidAlexandria
    Commented Sep 6, 2018 at 19:30
0
$\begingroup$

$y(x,t)=\sin(\omega\,t -k\, x)$ 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 $y(x,t)=\sin(\omega\,t -k\, x-\phi)$

Improve this answer
$\endgroup$
2
  • $\begingroup$ What should be the answer in my case? $\endgroup$
    – Yash Mittal
    Commented Sep 7, 2018 at 6:53
  • 1
    $\begingroup$ @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. $\endgroup$
    – rob ♦
    Commented Sep 8, 2018 at 18:42
0
$\begingroup$

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 $x-x_{0}$

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Why we can't put '-' sign infront of kx term? This will also give wave travelling in -ve direction. $\endgroup$
    – Yash Mittal
    Commented Sep 8, 2018 at 14:45
  • $\begingroup$ 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. $\endgroup$
    – Yash Mittal
    Commented Sep 8, 2018 at 14:52
0
$\begingroup$

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.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.