I am rather confused with the interference of waves that must occur in a string with different densities.
Say for example we have a string of length 2L. And the first L part has mass per unit length u, while the second part has mass per unit length 9u.
A wave is continuously propagated from the lighter string with the desired frequency.
Now the wave comes to the junction and some of it gets transmitted and some of it is reflected(W1) with phase difference $\pi$. The transmitted wave hits the other end and comes back with phase difference $\pi$ and again crosses the junction(W2).
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For standing waves to be formed does the W1 need to be in phase with the initial wave or does the W2 need to be in phase with the initial wave.
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Let's assume that I observe standing waves at a frequency $ f_1, f_2, f_3 … $ what would be the shape of the string. It can't be simple one loop,two loops, three loops respectively, as the wavelength of the wave changes when we go from one side to the other.
Best Answer
There is a lot of helpful information (and some great animations) on Daniel Russell's page with Acoustics and Vibration animations.
Of particular note is the fact that the amplitude of the reflected wave can be computed from the wave impedance. The wave impedance is given by
$$Z = \rho c = \sqrt{\rho T}$$
And the reflected wave amplitude is given by
$$A_r = \frac{Z_1 - Z_2}{Z_1+Z_2} A_i$$
When $Z_2 = 3Z_1$, it follows that $A_r = -\frac12 A_i$ for the wave traveling from left to right, and $A_r' = \frac12 A_i'$ for the wave traveling from right to left. The transmitted amplitude is given by
$$A_t = \frac{2Z_1}{Z_1+Z_2}$$
Traveling from low to high density, this is again $A_t = \frac12 A_i$, while from right to left it is $A_t' = \frac32 A_i'$.
Some of the bouncing of the waves is shown in this diagram:
At (a), a single pulse it traveling to the right. It partially reflects at the boundary, and a pulse of half the amplitude (and 1/3 of the wavelength) continues to the right, while the remainder is reflected and inverted at (b). At (c), the wave on the left has returned, while the one on the right is still traveling to the right. Another transmission/reflection happens, and you get an even smaller fraction of the wave on the left and a second pulse on the right at (d). If you continued this diagram, you would see that the motion on left and right is an infinite summation of waves of different amplitudes and timings; a steady state solution can only exist for certain frequencies, which we will compute below.
This diagram is what it would look like if you could give a short "kick" to the left end of the string, and watched the waves propagate. As these pulses travel back and forth, it will usually happen that the higher frequencies are damped and you are left with a standing wave. In principle you can do the same diagram with sine waves, but it would quickly look very mess - so let's go to the mathematical treatment instead:
It is known that the propagation velocity of the wave is proportional to the inverse square root of the mass per unit length; so if you have half the string at density $\rho$ and the other half at density $9\rho$, then the wave travels 3x faster in the thinner part of the string - and there should be more waves in the thick part.
To draw this, you need to find a function that is continuous in both amplitude (so string doesn't break) and the first derivative (otherwise there will be infinite acceleration at the "kink" until it's smooth again). This means that to the left of the center, it's of the form
$$y = A_1 \sin(k x)$$ while to the right it is
$$y = A_2 \sin(3k (2L-x))$$
(Using the $\sin$ basis function like this we enforce the boundary conditions at x=0 and x=2L).
Continuity of amplitude implies that
$$A_1 \sin(kL)=A_2\sin(3 kL)$$
and continuity of first derivative:
$$A_1 k \cos(kL)=-3 A_2 k \cos(3 kL)$$
Now we can solve for the wave number $k$ and the ratio of amplitudes in the two halves of the string. Putting $A_1=1$ for simplicity, we can divide the two equations by each other and find that
$$\tan{ kL} = -3 \tan(3 kL)$$
I am not smart enough to solve that equation - but Wolfram Alpha is. It gives me
$$\begin{align}kL &= n\pi\\ &= 2n\pi - 2\tan^{-1}\left(\sqrt{\frac13\left(13-4\sqrt{10}\right)}\right)\\ &= 2n\pi + 2\tan^{-1}\left(\sqrt{\frac13\left(13-4\sqrt{10}\right)}\right)\\ &= 2n\pi - 2\tan^{-1}\left(\sqrt{\frac13\left(13+4\sqrt{10}\right)}\right)\\ &= 2n\pi + 2\tan^{-1}\left(\sqrt{\frac13\left(13+4\sqrt{10}\right)}\right)\\ &\rm{with ~ n\in \mathbb{Z}}\end{align}$$
Here are plots of 4 different harmonics calculated from the above:
Interestingly, for the trivial mode where there is a node at the junction, the ratio of amplitudes of the waves needed for continuity is different (3:1 vs 1.5 : 1). Not sure why that is.
Disclaimer: it's possible there was a mistake in my math above... but I'm pretty sure that the principles are sound.