Applications of second order differential equations  Second Order Differential Equations
Applications of second order differential equations
Lessons
Notes:
Mechanical Vibrations:
Hooke's Law:
Force=ky
Newton's Second Law:
$Force= mass \times acceleration=m \times \frac{d^2y}{dt^2}$
$\Longrightarrow$ $m \times \frac{d^2y}{dt^2}=ky$ $\Longrightarrow$ $my'' + ky =0$Damping Force:
$Force = c\frac{dy}{dt}$ $\Longrightarrow$ $m \times \frac{d^2y}{dt^2}=c\frac{dy}{dt}ky$ $\Longrightarrow$ $my''+cy'+ky=0$Electrical Circuits:
Using second order differential equations we are able to analyze a circuit consisting of a battery, a resistor, an inductor and a capacitor in series. Let us denote $Q(t)$ as the charge on the capacitor at time $t$. The current is the rate of change of $Q$ with respect to $t$. So the current of the system is equal to $I=\frac{dQ}{dt}$
Kirchhoff's Law states that the sum of all voltage drops across a system must equal the supplied charge:
Where $V_L$ is the voltage drop across the inductor, $V_R$ is the voltage drop across the resistor, $V_C$ is the voltage drop across the capacitor, and $V_{bat}$ is the voltage supplied by the battery (or other electrical force).
Faraday's Law:
According to Faraday's Law the voltage drop across an inductor is equal to the instantaneous rate of change of current times an inductance constant, denoted by $L$ (measured in henry's).
Ohm's Law:
From Ohm's Law the voltage drop across a resistor is equal to the resistance (measured in ohms) times the current:
And the voltage drop across a capacitor is proportional to the electrical charge of the capacitor times a constant of capacitance (measured in farads).
And let us denote the voltage from the battery as some sort of function with respect to time $V_{bat}=E(t)$
So inputting all the previously found information into Kirchhoff's Law:
Which will become,
And we know that $I= \frac{dQ}{dt}$. So the equation becomes,
Which can also be written as
Which is a second order, constant coefficient, nonhomogeneous differential equation.

Intro Lesson

2.
Suppose that a hydraulic shock has a spring constant of 40 newtons per meter. There is a weight of 10kg attached to the end of the shock, and the shock has a resting length of 0.5 meters.