# Standing wave on a rope fixed at both sides: minus sign in the reflected wave

I'm studying stationary waves on a rope fixed at both sides. In some books I find that the wave function studied is the sum of incident wave $\xi_1(x,t)$ and of the reflected wave $\xi_2(x,t)$.

$$\xi(x,t)=\xi_1(x,t)+\xi_2(x,t)=A \mathrm{sin} (k x-\omega t)+ A \mathrm{sin}(kx+\omega t)=2 A \mathrm{sin}(kx)\mathrm{cos}(\omega t)\tag{1}$$

So this is the sum of two waves which differ only for the fact that one is progressive and one is regressive.

My doubts are on the fact that the fixed end of the rope cannot move, so there is a total reflection of $\xi_1(x,t)$ but the reflected wave $\xi_2(x,t)$ is in opposition of phase (i.e. uspide down), with respect to $\xi_1(x,t)$. So shouldn't $\xi_2(x,t)$ be

$$\xi_2(x,t)=- A \mathrm{sin}(kx+\omega t)$$

? The situation is the one in the picture. Then if this was correct, $(1)$ would change to

$$\xi(x,t)=\xi_1(x,t)+\xi_2(x,t)=A \mathrm{sin} (k x-\omega t)- A \mathrm{sin}(kx+\omega t)=2 A \mathrm{cos}(kx)\mathrm{sin}(\omega t)\tag{2}$$

Am I missing something or is the reasoning somehow correct? If so, are $(2)$ and $(1)$ equivalent?

They are not equivalent. A way to see this is to find the normal modes associated to these solutions.

Applying the condition $\xi(L,t)=0$ to $$\xi(x,t)=2 A \mathrm{sin}(kx)\mathrm{cos}(\omega t),$$ you get $kL=n\pi$. This gives $\lambda_n=2L/n$ and $f_n=nv/2L$. Which are the correct wavelength and frequency for the rope with both ends fixed.

For the second solution, $$\xi(x,t)=2 A \mathrm{cos}(kx)\mathrm{sin}(\omega t),$$ you obtain $kL=(2n-1)\pi/2$. The wavelength and frequency are $\lambda_n=4L/(2n-1)$ and $f_n=(2n-1)v/4L$, which are associated to the normal modes of a rope with one end fixed and one end free.

Hence the first solution is the correct one. On the other hand you are right when you say that the reflected wave must invert its phase. The solution of this apparent paradox is that the reflection is not simply $A\sin(kx-\omega t)\rightarrow -A(kx+\omega t)$. If you call $f(x-vt)$ the incident wave and $g(x+vt)$ the reflected wave, the complete solution is $$\xi(x,t)=f(x-vt)+g(x+vt).$$ Applying $\xi(0,t)=0$ we get $f(-vt)=-g(+vt)$ or $g(x')=-f(-x')$, $\forall x'$. In particular, this holds when $x'=x+vt$. The superposition reads $$\xi(x,t)=f(x-vt)-f(-(x+vt)).$$ If $f(x-vt)=A\sin(kx-\omega t)$, then $$\xi(x,t)=A\sin(kx-\omega t)+A\sin(kx+\omega t).$$

As we can see the correct reflection for a harmonic wave is $$A\sin(kx-\omega t)\rightarrow -A\sin(-(kx+\omega t))=A\sin(kx+\omega t).$$

Notice that if you had used $A\cos(kx\pm \omega t)$ instead you would get the reflection $$A\cos(kx-\omega t)\rightarrow -A\cos(kx+\omega t).$$

• @Sørën Please have another look at the answer. I tried to make the point you mentioned clear. May 14, 2016 at 23:53
• Thanks so much! It is clearer now, if I may ask, I still do not see very clearly how it is valid to go from $f(-vt)=-g(vt)$ to $g(x')=-f(-x')$ with $x'=x+vt$. I mean the condition $f(-vt)=-g(vt)$ is valid in $x=0$ (that's the boundary condition), then can we say from this that it is automatically valid for any other $x$, i.e. $g(x')=-f(-x')$ with $x'=x+vt$? May 15, 2016 at 6:16

If the phase difference between the wave is zero i.e lies in the plane of wave motion the resultant displacement is equal to zero, Thus, $A = 0$, $(L,t)=0$, due to that fact, you can use $$A\sin(kx−\omega t)\to-A\sin(-(kx+\omega t))=A\sin(kx+\omega t)$$ for progressive wave, but nothing can happen when you use cosine rule because give us answer equal to zero for the maximum nodes but $\sin Q = 1$.