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ContagiousKnowledge
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- Homework Statement
- Consider a thin elastic sheet that spreads from x = 0 to x = x' and y = 0 to y = y' where the sheet's transverse displacement ψ is 0 x=0, y=0, x=x', and y=y'. at Demonstrate that the amplitudes and phase angles in the normal mode expansion are given by $$ A_{i,j} = \sqrt {G^{2}{}_{i,j} + H^{2}{}_{i,j}} $$
and
$$ tan^{-1}(\frac {H_{i,j}} {G_{i,j}}) $$
Confirm that
$$ G_{i,j} = \frac {4} {x'y'} \int_{0}^{y'} \int_{0}^{x'} \psi(x, y, 0) sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) dx dy $$
$$ H_{i,j} = \frac {4} {x'y' \omega_{i,j}} \int_{0}^{y'} \int_{0}^{x'} \frac {\partial \psi(x, y, 0)} {\partial t} sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) dx dy $$
- Relevant Equations
- $$ \frac {\partial^2 \psi} {\partial t^2} = v^2( \frac {\partial^2 \psi} {\partial x^2} + \frac {\partial^2 \psi} {\partial y^2} ) $$
Let's try inputting a solution of the following form into the two-dimensional wave equation: $$ \psi(x, y, t) = X(x)Y(y)T(t) $$
Solving using the method of separation of variables yields
$$ \frac {v^2} {X(x)} \frac {\partial^2 X(x)} {\partial x^2} + \frac {v^2} {Y(y)} \frac {\partial^2 Y(y)} {\partial y^2} =\frac {1} {T(t)} \frac {\partial^2 T(t)} {\partial t^2} =-\omega^2$$
$$ T(t) = A_{i,j}cos(\omega_{i,j} t) + B_{i,j}sin(\omega_{i,j} t)$$
$$ X(x) =Ccos(k_{x} t) + D_{i,j}sin(k_{x} t) $$
$$ Y(y) =Ecos(k_{y} t) + Fsin(k_{y} t) $$
where
$$ \frac {1} {X(x)} \frac {\partial^2 X(x)} {\partial x^2} = -k_{x}^2 $$
$$ \frac {1} {Y(y)} \frac {\partial^2 Y(y)} {\partial y^2} = -k_{y}^2 $$
The boundary conditions ψ(0, y, t) = ψ(x, 0, t) = 0 tells us that C=E=0.
the conditions ψ(x', y, t) = ψ(x, y', t) = 0 tell us that
$$ k_{x} = \frac {\pi i} {x'}, $$
$$ k_{y} = \frac {\pi j} {y'} $$
$$ \omega = \pi v \sqrt {\frac{i^{2}}{x'^{2}}+\frac{j^{2}}{y'^{2}}} $$
From this, we can ascertain the particular solution for given i and j
$$ \psi(x, y, t) = [G_{i,j}cos(\omega_{i,j} t) + H_{i,j}sin(\omega_{i,j} t)]sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y)$$
$$ G_{i,j}=A_{i,j}D_{i}F_{j}, H_{i,j}=B_{i,j}D_{i}F_{j} $$
Since the wave equation is linear, the general solution is a superposition of all normal modes.
$$ \psi(x, y, t) = \Sigma_{i=0,\infty} \Sigma_{j=0,\infty}[G_{i,j}cos(\omega_{i,j} t) + H_{i,j}sin(\omega_{i,j} t)]sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) $$
$$ \psi(x, y, 0) = \Sigma_{i=0,\infty} \Sigma_{j=0,\infty}G_{i,j}sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) $$
$$ G_{i,j} = \frac {4} {x'y'} \int_{0}^{y'} \int_{0}^{x'} \psi(x, y, 0) sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) dx dy $$
$$ \frac {\partial \psi(x, y, 0)} {\partial t} = \Sigma_{i=0,\infty} \Sigma_{j=0,\infty} \omega_{i,j} H_{i,j}sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) $$
$$ H_{i,j} = \frac {4} {x'y' \omega_{i,j}} \int_{0}^{y'} \int_{0}^{x'} \frac {\partial \psi(x, y, 0)} {\partial t} sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) dx dy $$
I've omitted some math because entering it with latex was taking a long time, but I have arrived at the correct expressions for G and H. I don't know how to proceed from here.
Solving using the method of separation of variables yields
$$ \frac {v^2} {X(x)} \frac {\partial^2 X(x)} {\partial x^2} + \frac {v^2} {Y(y)} \frac {\partial^2 Y(y)} {\partial y^2} =\frac {1} {T(t)} \frac {\partial^2 T(t)} {\partial t^2} =-\omega^2$$
$$ T(t) = A_{i,j}cos(\omega_{i,j} t) + B_{i,j}sin(\omega_{i,j} t)$$
$$ X(x) =Ccos(k_{x} t) + D_{i,j}sin(k_{x} t) $$
$$ Y(y) =Ecos(k_{y} t) + Fsin(k_{y} t) $$
where
$$ \frac {1} {X(x)} \frac {\partial^2 X(x)} {\partial x^2} = -k_{x}^2 $$
$$ \frac {1} {Y(y)} \frac {\partial^2 Y(y)} {\partial y^2} = -k_{y}^2 $$
The boundary conditions ψ(0, y, t) = ψ(x, 0, t) = 0 tells us that C=E=0.
the conditions ψ(x', y, t) = ψ(x, y', t) = 0 tell us that
$$ k_{x} = \frac {\pi i} {x'}, $$
$$ k_{y} = \frac {\pi j} {y'} $$
$$ \omega = \pi v \sqrt {\frac{i^{2}}{x'^{2}}+\frac{j^{2}}{y'^{2}}} $$
From this, we can ascertain the particular solution for given i and j
$$ \psi(x, y, t) = [G_{i,j}cos(\omega_{i,j} t) + H_{i,j}sin(\omega_{i,j} t)]sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y)$$
$$ G_{i,j}=A_{i,j}D_{i}F_{j}, H_{i,j}=B_{i,j}D_{i}F_{j} $$
Since the wave equation is linear, the general solution is a superposition of all normal modes.
$$ \psi(x, y, t) = \Sigma_{i=0,\infty} \Sigma_{j=0,\infty}[G_{i,j}cos(\omega_{i,j} t) + H_{i,j}sin(\omega_{i,j} t)]sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) $$
$$ \psi(x, y, 0) = \Sigma_{i=0,\infty} \Sigma_{j=0,\infty}G_{i,j}sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) $$
$$ G_{i,j} = \frac {4} {x'y'} \int_{0}^{y'} \int_{0}^{x'} \psi(x, y, 0) sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) dx dy $$
$$ \frac {\partial \psi(x, y, 0)} {\partial t} = \Sigma_{i=0,\infty} \Sigma_{j=0,\infty} \omega_{i,j} H_{i,j}sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) $$
$$ H_{i,j} = \frac {4} {x'y' \omega_{i,j}} \int_{0}^{y'} \int_{0}^{x'} \frac {\partial \psi(x, y, 0)} {\partial t} sin(\frac {\pi i} {x'}x)sin(\frac {\pi j} {y'}y) dx dy $$
I've omitted some math because entering it with latex was taking a long time, but I have arrived at the correct expressions for G and H. I don't know how to proceed from here.
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