Vibración no amortiguada con carga impulsiva rectangular

\begin{equation*} \ddot{u} + \omega^{2} u = \begin{cases} \frac{F_{0}}{m} & 0 \leqslant t \leqslant t_{1} \\ 0 & t \geqslant 0 \end{cases} \end{equation*}

La solución es

\begin{equation*} u = \begin{cases} A_{1} \sin \bigl( \omega t \bigr) + B_{1} \cos \bigl( \omega t \bigr) + \frac{F_{0}}{k} & 0 \leqslant t \leqslant t_{1} \\ A_{2} \sin \bigl( \omega t \bigr) + B_{2} \cos \bigl( \omega t \bigr) & t \geqslant 0 \end{cases} \end{equation*}

Coeficientes

\begin{align*} A_{1} &= \frac{\dot{u}_{0}}{\omega} \\ B_{1} &= u_{0} - \frac{F_{0}}{k} \\ A_{2} &= \frac{\dot{u}(t_{1})}{\omega} \\ B_{2} &= u(t_{1}) \end{align*}

Ejemplo en clase

Usando una traslación de coordenadas en la vibración libre

\begin{equation*} u = \begin{cases} A_{1} \sin \bigl( \omega t \bigr) + B_{1} \cos \bigl( \omega t \bigr) + \frac{F_{0}}{k} & 0 \leqslant t \leqslant t_{1} \\ A_{2} \sin \bigl[ \omega \bigl( t - t_{1} \bigr) \bigr] + B_{2} \cos \bigl[ \omega \bigl( t - t_{1} \bigr) \bigr] & t \geqslant t_{1} \end{cases} \end{equation*}

In [1]:
import numpy as np
import matplotlib.pyplot as plt

w = 42.0    # Tn
m = w/981.0 # Tn/cm/seg^2
k = 10.5    # Tn/cm
u0 = 0.0    # cm
v0 = 0.0    # cm/seg
F0 = 2.0    # Tn
t1 = 1.0    # segundos

omega = np.power(k/m,1.0/2.0)
A1 = v0/omega
B1 = u0 - F0/k
A2 = A1*np.cos(omega*t1) - B1*np.sin(omega*t1)
B2 = A1*np.sin(omega*t1) + B1*np.cos(omega*t1) + F0/k

print 'omega =', omega
print 'A1 =', A1
print 'B1 =', B1
print 'A2 =', A2
print 'B2 =', B2

omega = 15.6604597634
A1 = 0.0
B1 = -0.190476190476
A2 = 0.00904488393634
B2 = 0.380737508702
In [2]:
t = np.linspace(0,3,3000)
posicion = np.where(t<=t1, A1*np.sin(omega*t) + B1*np.cos(omega*t) + F0/k, A2*np.sin(omega*(t-t1)) + B2*np.cos(omega*(t-t1)))

plt.figure(figsize=(19,8.5))
plt.plot(t,posicion)
plt.xlabel(r'$t$  seg.')
plt.ylabel(r'$u$  cm.')
plt.grid(True)
plt.show()

Cuando $u_{0} = 0$ y $\dot{u}_{0} = 0$

\begin{equation*} u = \begin{cases} B_{1} \cos \bigl( \omega t \bigr) + \frac{F_{0}}{k} & 0 \leqslant t \leqslant t_{1} \\ A_{2} \sin \bigl[ \omega \bigl( t - t_{1} \bigr) \bigr] + B_{2} \cos \bigl[ \omega \bigl( t - t_{1} \bigr) \bigr] & t \geqslant t_{1} \end{cases} \end{equation*}

Coeficientes

\begin{align*} B_{1} &= - \frac{F_{0}}{k} \\ A_{2} &= \frac{\dot{u}(t_{1})}{\omega} \\ B_{2} &= u(t_{1}) \end{align*}

In [3]:
import numpy as np
import matplotlib.pyplot as plt

w = 42.0    # Tn
m = w/981.0 # Tn/cm/seg^2
k = 10.5    # Tn/cm
F0 = 2.0    # Tn
t1 = 1.0    # segundos

omega = np.power(k/m,1.0/2.0)
B1 = -F0/k
A2 = -B1*np.sin(omega*t1)
B2 = B1*np.cos(omega*t1) + F0/k

print 'omega =', omega
print 'B1 =', B1
print 'A2 =', A2
print 'B2 =', B2

omega = 15.6604597634
B1 = -0.190476190476
A2 = 0.00904488393634
B2 = 0.380737508702
In [4]:
t = np.linspace(0,3,3000)
posicion = np.where(t<=t1, B1*np.cos(omega*t) + F0/k, A2*np.sin(omega*(t-t1)) + B2*np.cos(omega*(t-t1)))

plt.figure(figsize=(19,8.5))
plt.plot(t,posicion)
plt.xlabel(r'$t$  seg.')
plt.ylabel(r'$u$  cm.')
plt.grid(True)
plt.show()

Usando odeint

\begin{align*} u^{\prime} &= z \\ z^{\prime} + \omega^{2} u &= \begin{cases} \frac{F_{0}}{m} & 0 \leqslant t \leqslant 1 \\\ 0 & t \geqslant 1 \end{cases} \end{align*}

sujeto a $u(0) = 0$ y $u^{\prime}(0) = 0$

In [5]:
from scipy.integrate import odeint
import numpy as np
import matplotlib.pyplot as plt

def dU_dx(U, x):
    # Here U is a vector such that y=U[0] and z=U[1]. This function should return [z', u']
    m = 42.0/981.0
    k = 10.5
    omega = np.power(k/m,1.0/2.0)
    F0 = 2.0
    t1 = 1.0 # segundos
    if x <= t1:
        return [U[1],-omega**2*U[0] + F0/m]
    else:
        return [U[1],-omega**2*U[0]]
U0 = [0.0,0.0]
xs = np.linspace(0, 3, 3000)
Us = odeint(dU_dx, U0, xs)
ys = Us[:,0]

plt.figure(figsize=(19,8.5))
plt.plot(xs, ys)
plt.xlabel(r'$t$  seg.')
plt.ylabel(r'$u$  cm.')
plt.grid(True)
plt.show()

Periodo

\begin{equation*} T = \frac{2 \pi}{\omega} \end{equation*}

In [6]:
T = (2.0*np.pi)/omega
print 'T =', T

T = 0.401213336142
In [7]:
valor_pico = np.where(posicion >= max(posicion))[0]
print 't pico =', t[valor_pico[0]]
print 'valor pico =', posicion[valor_pico[0]]

t pico = 0.602200733578
valor pico = 0.380948995231

Gráfico interactivo

In [8]:
import plotly.offline as offline
import plotly.graph_objs as go
import numpy as np

offline.init_notebook_mode()

t = np.linspace(0,3,3000)
posicion = np.where(t<=t1, A1*np.sin(omega*t) + B1*np.cos(omega*t) + F0/k, A2*np.sin(omega*(t-t1)) + B2*np.cos(omega*(t-t1)))
trace = go.Scatter(x = t, y = posicion)
data = [trace]

offline.iplot(data)

Desplazamientos máximos

Usando la transformada de Hilbert

In [9]:
import numpy as np
import matplotlib.pyplot as plt
from scipy.signal import hilbert

t = np.linspace(0,3,3000)
posicion = np.where(t<=t1, A1*np.sin(omega*t) + B1*np.cos(omega*t) + F0/k, A2*np.sin(omega*(t-t1)) + B2*np.cos(omega*(t-t1)))
analytic_signal = hilbert(posicion)
amplitude_envelope = np.abs(analytic_signal)
instantaneous_phase = np.unwrap(np.angle(analytic_signal))
instantaneous_frequency = (np.diff(instantaneous_phase)/(2.0*np.pi))
fig = plt.figure(figsize=(19,8.5))
ax0 = fig.add_subplot(211)
ax0.plot(t, posicion, label='posicion')
ax0.plot(t, amplitude_envelope, label='envolvente')
ax0.set_xlabel("tiempo en segundos")
ax0.legend()
ax1 = fig.add_subplot(212)
ax1.plot(t[1:], instantaneous_frequency)
ax1.set_xlabel("tiempo en segundos")

plt.show()

Usando argrelextrema

In [10]:
import numpy as np
from scipy.signal import argrelextrema
import matplotlib.pyplot as plt

t = np.linspace(0,3,3000)
posicion = np.where(t<=t1, A1*np.sin(omega*t) + B1*np.cos(omega*t) + F0/k, A2*np.sin(omega*(t-t1)) + B2*np.cos(omega*(t-t1)))

maximo = argrelextrema(posicion, np.greater)[0] #array of indexes of the locals maxima
x = [t[maximo] for i in maximo][0]
y = [posicion[maximo] for i in maximo][0]

plt.figure(figsize=(19,8.5))
plt.plot(t,posicion)
plt.plot(x, y, 'ro')
plt.xlabel(r'$t$  seg.')
plt.ylabel(r'$u$  cm.')
plt.show()

Desplazamiento máximo

In [11]:
for i in range(np.size(x)):
    print i, '\t', round(x[i],3), '\t', round(y[i],3)

0 	0.201 	0.381
1 	0.602 	0.381
2 	1.001 	0.381
3 	1.402 	0.381
4 	1.804 	0.381
5 	2.205 	0.381
6 	2.607 	0.381

Periodo aproximado

In [12]:
for i in range(np.size(x) - 1):
    print i+1, '\t', x[i+1]-x[i]

1 	0.401133711237
2 	0.399133044348
3 	0.401133711237
4 	0.401133711237
5 	0.401133711237
6 	0.402134044682

Espectro de respuesta

Desplazamientos

\begin{equation*} u = \begin{cases} \frac{F_{0}}{k} \bigl[ 1 - \cos \bigl( \omega t \bigr) \bigr] & 0 \leqslant t \leqslant t_{1} \\ A_{2} \sin \bigl( \omega t \bigr) + B_{2} \cos \bigl( \omega t \bigr) & t \geqslant 0 \end{cases} \end{equation*}

Valores conocidos

\begin{align*} \omega t &= 2 \pi \frac{t}{T} \\ \sqrt{A_{2}^{2} + B_{2}^{2}} &= \frac{F_{0}}{k} \sqrt{\bigl[ \sin \bigl( \omega t \bigr) \bigr]^{2} + \bigl[ 1 - \cos \bigl( \omega t \bigr) \bigr]^{2}} = \frac{F_{0}}{k} \sqrt{2 \bigl[ 1 - \cos \bigl( \omega t \bigr) \bigr]} = \frac{F_{0}}{k} \big | 2 \sin \bigl( \frac{\omega t}{2} \bigr) \big | \end{align*}

Reemplazando

\begin{equation*} u = \begin{cases} \frac{F_{0}}{k} \bigl[ 1 - \cos \bigl( 2 \pi \frac{t}{T} \bigr) \bigr] & 0 \leqslant t \leqslant t_{1} & \mbox{respuesta forzada} \\ \frac{2 F_{0}}{k} & t \geqslant t_{1} & \mbox{respuesta forzada} \\ \frac{2 F_{0}}{k} \big | \sin \bigl( \pi \frac{t}{T} \bigr) \big | & t \geqslant 0 & \mbox{respuesta libre} \end{cases} \end{equation*}

Reescribiendo

\begin{equation*} u = \begin{cases} \frac{F_{0}}{k} \bigl[ 1 - \cos \bigl( 2 \pi \frac{t}{t_{1}} \frac{t_{1}}{T} \bigr) \bigr] & 0 \leqslant \frac{t_{1}}{T} \leqslant \frac{1}{2} & \mbox{respuesta forzada} \\ \frac{2 F_{0}}{k} & \frac{t_{1}}{T} \geqslant \frac{1}{2} & \mbox{respuesta forzada} \\ \frac{2 F_{0}}{k} \big | \sin \bigl( \pi \frac{t}{t_{1}} \frac{t_{1}}{T} \bigr) \big | & \frac{t_{1}}{T} \geqslant 0 & \mbox{respuesta libre} \end{cases} \end{equation*}

Para simplificar $\frac{t}{t_{1}} = 1$

In [13]:
t = np.linspace(0,3,3000)
u0 = 2*np.abs(np.sin(np.pi*t))
u1 = np.where(t<=0.5, 1 - np.cos(2*np.pi*t), 2)
plt.figure(figsize=(11,8.5))
plt.plot(t,u0, '--', label='respuesta libre')
plt.plot(t,u1, label='respuesta forzada')
plt.legend(loc='upper right')
plt.xlabel(r'$\frac{t_{1}}{T}$')
plt.ylabel(r'$R_{d}$')
plt.grid(True)
plt.show()

El espectro de respuesta es la envolvente

\begin{equation*} u_{\max} = \begin{cases} \frac{2 F_{0}}{k} \big | \sin \bigl( \pi \frac{t_{1}}{T} \bigr) \big | & 0 \leqslant \frac{t_{1}}{T} \leqslant \frac{1}{2} & \mbox{respuesta libre} \\ \frac{2 F_{0}}{k} & \frac{t_{1}}{T} \geqslant \frac{1}{2} & \mbox{respuesta forzada} \end{cases} \end{equation*}

In [14]:
t = np.linspace(0,3,3000)
u = np.where(t<=0.5, 2*np.abs(np.sin(np.pi*t)), 2)

plt.figure(figsize=(11,8.5))
plt.plot(t,u)
plt.xlabel(r'$\frac{t_{1}}{T}$')
plt.ylabel(r'$R_{d}$')
plt.grid(True)
plt.show()

Factor de amplificación máximo

In [15]:
valor_pico = np.where(u>= max(u))[0]
print 't1/T =', 1/(2.0*t[valor_pico[0]])
print 'Rd =', u[valor_pico[0]]

t1/T = 0.999666666667
Rd = 2.0