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Commit b149738c authored by Erik Strand's avatar Erik Strand
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Add something for 3.1 (d)

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from sympy import *
from sympy.assumptions.assume import global_assumptions
#from sympy.abc import x, y
#expr = spy.sin(x) / spy.cos(x)
#spy.pprint(expr)
#expr = spy.simplify(expr)
#spy.pprint(expr)
alpha, omega_1, t, A, B = symbols('alpha omega_1 t A B')
x = exp(-alpha * t) * (A * exp(I * omega_1 * t) + B * exp(-I * omega_1 * t))
dx = simplify(diff(x, t))
ddx = simplify(diff(dx, t))
pprint(x)
print("")
pprint(dx)
print("")
pprint(ddx)
m, k = symbols('m k')
global_assumptions.add(Q.real(m))
global_assumptions.add(Q.positive(m))
global_assumptions.add(Q.real(k))
global_assumptions.add(Q.positive(k))
omega_0, gamma = symbols('omega_0 gamma')
E = simplify(Rational(1, 2) * (m * dx**2 + k * x**2))
E_0 = simplify(E.subs(t, 0))
#E_0 = simplify(E.subs([(t, 0), (omega_1, sqrt(omega_0**2 - alpha**2))]))
#E_0 = simplify(E_0.subs([(omega_0, sqrt(k / m)), (alpha, gamma / (2 * m))]))
print("")
pprint(E)
print("")
pprint(E_0)
dE = -gamma * dx
delta_E = simplify(integrate(dE, (t, 0, 2 * pi / omega_1)))
print("")
pprint(delta_E)
Q = 2 * pi * E_0 / delta_E
Q = Q.subs(omega_1, sqrt(omega_0**2 - alpha**2))
Q = Q.subs([
(omega_0, sqrt(k / m)),
(alpha, gamma / (2 * m))
])
Q = simplify(Q)
print("")
pprint(Q)
...@@ -134,6 +134,17 @@ $$ ...@@ -134,6 +134,17 @@ $$
A = \frac{1}{-m \omega^2 + i \gamma \omega + k} A = \frac{1}{-m \omega^2 + i \gamma \omega + k}
$$ $$
Thus the general solution to the inhomogeneous problem (assuming distinct roots) is
$$
x(t) = e^{-\frac{\gamma}{2m} t} \left(
A e^{\frac{\sqrt{\gamma^2 - 4mk}}{2m} t} +
B e^{-\frac{\sqrt{\gamma^2 - 4mk}}{2m} t}
\right) + \frac{1}{-m \omega^2 + i \gamma \omega + k} e^{i \omega t}
$$
Here are plots of the magnitude and phase of the particular solution.
![amplitude](../assets/img/02_amplitude.png) ![amplitude](../assets/img/02_amplitude.png)
![phase](../assets/img/02_phase.png) ![phase](../assets/img/02_phase.png)
...@@ -149,12 +160,126 @@ to the energy lost per radian (one cycle is $$2\pi$$ radians). Show that these t ...@@ -149,12 +160,126 @@ to the energy lost per radian (one cycle is $$2\pi$$ radians). Show that these t
equal, assuming that the damping is small. How long does it take the amplitude of a 100 Hz equal, assuming that the damping is small. How long does it take the amplitude of a 100 Hz
oscillator with a Q of 109 to decay by $$1/e$$? oscillator with a Q of 109 to decay by $$1/e$$?
#### Undamped
As we found previously, the general solution is
$$
x(t) = e^{-\frac{\gamma}{2m} t} \left( A e^{\frac{\sqrt{\gamma^2 - 4mk}}{2m} t} + B e^{-\frac{\sqrt{\gamma^2 - 4mk}}{2m} t} \right)
$$
To get rid of a symbol, let's define $$\omega_0 = \sqrt{k/m}$$ and $$\alpha = \gamma / (2 m)$$.
Then
$$
\frac{\sqrt{\gamma^2 - 4mk}}{2m} = \sqrt{\alpha^2 - \omega_0^2}
$$
Since we're assuming low damping, $$\alpha^2 - \omega_0^2$$ is negative. So we'll use
$$
\sqrt{\alpha^2 - \omega_0^2} = i \sqrt{\omega_0^2 - \alpha^2}
$$
Finally, to be as lazy as possible, we'll define $$\omega_1^2 = \omega_0^2 - \alpha^2$$ (which
is a positive real number). All together this compresses our general solution to
$$
\begin{aligned}
x(t)
&= e^{-\alpha t} \left( A e^{\sqrt{\alpha^2 - \omega_0^2} t} + B e^{-\sqrt{\alpha^2 - \omega_0^2} t} \right) \\
&= e^{-\alpha t} \left( A e^{i \sqrt{\omega_0^2 - \alpha^2} t} + B e^{-i \sqrt{\omega_0^2 - \alpha^2} t} \right) \\
&= e^{-\alpha t} \left( A e^{i \omega_1 t} + B e^{-i \omega_1 t} \right) \\
\end{aligned}
$$
The energy in the system is
$$
E = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} k x^2
$$
It's easy enough to see that
$$
x^2(t) = e^{-2 \alpha t} \left( A e^{i \omega_1 t} + B e^{-i \omega_1 t} \right)^2
$$
Applying the product rule to the general form as written above, and doing some tedious bookkeeping,
we can compute the $$\dot{x}$$ terms we need for the energy of the system.
$$
\begin{aligned}
\dot{x}(t) &=
-\alpha e^{-\alpha t} \left( A e^{i \omega_1 t} + B e^{-i \omega_1 t} \right)
+ i \omega_1 e^{-\alpha t} \left( A e^{i \omega_1 t} - B e^{-i \omega_1 t} \right) \\
\dot{x}(t)^2 &=
e^{-2 \alpha t} \big[ \alpha^2 \left( A e^{i \omega_1 t} + B e^{-i \omega_1 t} \right)^2 \\
&- i \alpha \omega_1 \left( A e^{i \omega_1 t} + B e^{-i \omega_1 t} \right) \left( A e^{i \omega_1 t} - B e^{-i \omega_1 t} \right) \\
&- \omega_1^2 \left( A e^{i \omega_1 t} - B e^{-i \omega_1 t} \right)^2 \big]
\end{aligned}
$$
So at time $$t = 0$$ the energy in the system is
$$
\begin{aligned}
E &= \frac{1}{2} m \big[ \alpha^2 (A + B)^2 - i \alpha \omega_1 (A + B)(A - B) - \omega_1^2 (A - B)^2 \big] \\
&+ \frac{1}{2} k (A + B)^2 \\
&= \frac{1}{2} \big[ (m \alpha^2 + k) (A + B)^2 - i m \alpha \omega_1 (A + B)(A - B) - m \omega_1^2 (A - B)^2 \big]
\end{aligned}
$$
To compute the change in the energy of the system, we'd like to integrate $$dE/dt$$. Note that
$$
\begin{aligned}
E &= \frac{1}{2} m \dot{x}^2 + \frac{1}{2} k x^2 \\
\frac{dE}{dt} &= m \dot{x} \ddot{x} + k x \dot{x} \\
&= \dot{x} (m \ddot{x} + k x) \\
&= -\gamma \dot{x}^2
\end{aligned}
$$
The last line follows because the ODE governing the system says $$m \ddot{x} + k x = -\gamma
\dot{x}$$.
...and I got caught in a sympy rabbit hole. Think it'll be valuable for future problem sets but
cost me too much time on this one.
### (e) ### (e)
{:.question} {:.question}
Now find the solution to equation (3.58) by using Laplace transforms. Take the initial condition as Now find the solution to equation (3.58) by using Laplace transforms. Take the initial condition as
$$x(0) = \dot{x}(0) = 0$$. $$x(0) = \dot{x}(0) = 0$$.
Using the table in the book, and keeping in mind the initial conditions, I found the following
transforms.
$$
\begin{aligned}
\mathcal{L}[m \ddot{x}(t)]
&= m \left( s^2 X(s) - s x(0) - \dot{x}(0) \right) \\
&= m s^2 X(s) \\
\mathcal{L}[\gamma \dot{x}(t)]
&= \gamma \left( s X(s) - x(0) \right) \\
&= \gamma s X(s) \\
\mathcal{L}[k x(t)]
&= k X(s) \\
\mathcal{L}[e^{i \omega t}] &= \frac{1}{s - i \omega}
\end{aligned}
$$
So our transformed ODE is
$$
m s^2 X(s) + \gamma s X(s) + k X(x) = \frac{1}{s - i \omega} \\
X(s) = \frac{1}{(s - i \omega) (m s^2 + \gamma s + k)}
$$
Now it's just a bunch of algebra and an inverse transform...
### (f) ### (f)
{:.question} {:.question}
......
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