- #1
cryora
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Homework Statement
I'm not exactly where a question like this would belong, so I am going to try here. It feels like as we go higher in our courses, we're expected to be able to work with more complicated equations. One example would be a problem involving transmission reflection ratios for waves:
An infinite continuous string with tension, τ, and at the ends has a point mass, m, attached to it at x=0. The mass is attached to a spring with natural frequency ω0 and is at equilibrium when the string is straight. A wave is incident from the left. Find the transmission and reflection ratios.
(There is no gravity in this problem)
Homework Equations
I already know the solution, but it starts with writing the Lagrangian by first assuming the string is made up of discrete masses (q0 being the generalized coordinate for the transverse displacement of mass m from equilibrium):
$$L=T-V$$
$$
L=\sum_{j\neq 0} \frac{1}{2} \mu \dot{q_j}^2 + \frac{1}{2} m \dot{q_0}^2 - \frac{\tau}{2d} \sum_j (q_{j+1} - q_{j})^2 - \frac{1}{2}m\omega_0^2q_0^2
$$
Then taking the Euler-Lagrange for q0 (and adding a damping term -bq0' since nonconservative forces cannot be derived from potentials):
$$
\frac{\partial L}{\partial q_0} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q_0}}=-m\omega_0^2q_0 - \tau (\frac{q_{0} - q_{-1}}{d}-\frac{q_{1} - q_{0}}{d})-m\ddot{q_0}-b\dot{q_0}=0
$$
Then taking the limit as d->0, we have a continuous wave equation:
$$
m\ddot{\psi_0}=-m\omega_0^2\psi_0 - \tau (\frac{\partial \psi_L}{\partial x}|_{x=0} - \frac{\partial \psi_R}{\partial x}|_{x=0})-m\ddot{\psi_0}-b\dot{\psi_0}
$$
Where the psi's are determined by the traveling wave ansatz and boundary conditions.
Wavefunction from the left:
$$
\psi_L(x,t)=A e^{i(\omega t - k x)} + B e^{i(\omega t + k x)}
$$
Wavefunction to the right:
$$
\psi_R(x,t)=C e^{i(\omega t - k x)}
$$
Wavefunction at x=0:
$$
\psi_0(t)≡\psi_R(0,t)=C e^{i\omega t}
$$
With the boundary condition that psi is continuous at x = 0, we have:
$$
C = A + B
$$
Plugging these values into the wave equation, we can solve for C/A and B/A so that
$$
T=\frac{|C|^2}{|A|^2}
$$
$$
R=\frac{|B|^2}{|A|^2}
$$
The Attempt at a Solution
When I start solving for the ratios is where the really complicated expressions begin. According to the TA's notes, B/A should be:
$$
\frac{B}{A}=\frac{-b \omega (2k\tau+b\omega)-m^2(\omega^2-\omega_0^2)^2-2ik\tau m(\omega^2-\omega_0^2)}{(2k\tau+b\omega)^2+m^2(\omega^2-\omega_0^2)^2}
$$
So that
$$
R=\frac{\frac{b^2\omega^2}{4k^2\tau^2}+\frac{m^2}{4k^2\tau^2}(\omega^2-\omega_0^2)^2}{(1+\frac{b\omega}{2k\tau})^2+\frac{m^2}{4k^2\tau^2}(\omega^2-\omega_0^2)^2}
$$
While I was able to get the same expression for B/A, I got a different expression for R. I basically squared the real part and the imaginary part of the numerator and added them, and squared the denominator:
(My Attempt by Hand)
$$
R=\frac{[-m^2(\omega^2-\omega_0^2)^2-b\omega(b\omega+2\tau k)]^2+4\tau^2k^2m^2(\omega^2-\omega_0^2)^2}{[m^2(\omega^2-\omega_0^2)^2+(b\omega+2\tau k)^2]^2}
$$
This expression doesn't look like it can be reconciled to look like my TA's R.
However, when doing it on Mathematica, I get something that is equivalent to my TA's R:
(Result from using Mathematica)
$$
R=1-\frac{4k\tau (k\tau + b\omega)}{4k^2 \tau^2 + 4bk\tau \omega + b^2\omega^2+m^2\omega^4+m^2\omega_0^2(-2\omega^2+\omega_0^2)}
$$
Working with these complicated expressions is really time consuming, and I want to gain some insight on how TA's and Professors solve these problems. Is it just a matter of practice and knowing how the solution is going to turn out? Do I have to be proficient and making more calculations in my head? Are there tricks like substituting big expressions with variables? Should I start using technology more, rather than doing everything by hand on lined paper? Should I use blank white paper?
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