Applications of second order differential equations

Everything You Need in One Place

Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

Learn and Practice With Ease

Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

Instant and Unlimited Help

Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!

Get the most by viewing this topic in your current grade. Pick your course now.

?
Intros
Lessons
  2. What are some applications of second order differential equations?
  3. Mechanical Vibrations and Dampening Forces
  4. Damping Forces on Mechanical Vibrations
  5. Electrical Circuits
?
Examples
Lessons

  1. Mechanical Vibrations

    A spring has a weight of 5kg attached to the end of it. The spring has a natural length of 0.3m and a force of 35 newtons is required to stretch the spring to a length 0.8m. If the spring is stretched to 0.5 meters and then released (with zero initial velocity), then what is the position of the mass at time tt?
    1. 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.
      1. What is the position of the mass at time tt if the hydraulic shock has a damping constant of c=50c=50, with an initial positions of 0.75 meters, and an initial velocity of 0 m/sm/s?
      2. What is the position of the mass at time tt if the hydraulic shock has a damping constant of c=40c=40, with an initial position of 0.5 meters, and an initial velocity of 5 m/sm/s?
      3. What is the position of the mass at time tt if the hydraulic shock has a damping constant of c=20c=20, with an initial position of 0.3 meters, and an initial velocity of -0.3 m/sm/s?

    2. Electrical Circuits

      Find the charge at time tt for an electrical circuit with a resistor that has a resistance of R=14ΩR=14 \Omega , an inductor with L=2HL=2H, a capacitor with C=0.05FC=0.05 F, and a battery with charge E(t)=8E(t)=8sin⁡(2t)\sin(2t). The initial charge is V=2229V=\frac{22}{29} coulombs, and the initial current is I=629I=\frac{6}{29} amps.

      Topic Notes
      ?

      Mechanical Vibrations:

      Hooke's Law:

      Force=-ky

      Newton's Second Law:

      Force=mass×acceleration=m×d2ydt2Force= mass \times acceleration=m \times \frac{d^2y}{dt^2}

      ⟹\Longrightarrow m×d2ydt2=−kym \times \frac{d^2y}{dt^2}=-ky

      ⟹\Longrightarrow my′′+ky=0my'' + ky =0

      Damping Force:

      Force=−cdydtForce = -c\frac{dy}{dt}

      ⟹\Longrightarrow m×d2ydt2=−cdydt−kym \times \frac{d^2y}{dt^2}=-c\frac{dy}{dt}-ky

      ⟹\Longrightarrow my′′+cy′+ky=0my''+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)Q(t) as the charge on the capacitor at time tt. The current is the rate of change of QQ with respect to tt. So the current of the system is equal to I=dQdtI=\frac{dQ}{dt}

      Kirchhoff's Law states that the sum of all voltage drops across a system must equal the supplied charge:

      VL+VR+VC=VbatV_L+V_R+V_C=V_{bat}

      Where VLV_L is the voltage drop across the inductor, VRV_R is the voltage drop across the resistor, VCV_C is the voltage drop across the capacitor, and VbatV_{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 LL (measured in henry's).

      VL=L×dIdtV_L=L \times \frac{dI}{dt}

      Ohm's Law:

      From Ohm's Law the voltage drop across a resistor is equal to the resistance (measured in ohms) times the current:

      VR=IRV_R=IR

      And the voltage drop across a capacitor is proportional to the electrical charge of the capacitor times a constant of capacitance (measured in farads).

      VC=1C×Q(t)V_C=\frac{1}{C} \times Q(t)

      And let us denote the voltage from the battery as some sort of function with respect to time Vbat=E(t)V_{bat}=E(t)

      So inputting all the previously found information into Kirchhoff's Law:

      VL+VR+VC=VbatV_L+V_R+V_C=V_{bat}

      Which will become,

      LdIdt+IR+1CQ(t)=E(t)L \frac{dI}{dt}+IR+\frac{1}{C} Q(t)=E(t)

      And we know that I=dQdtI= \frac{dQ}{dt}. So the equation becomes,

      Ld2Qdt2+RdQdt+1CQ(t)=E(t)L \frac{d^2 Q}{dt^2}+R \frac{dQ}{dt}+\frac{1}{C} Q(t)=E(t)

      Which can also be written as

      LQ′′+RQ′+1CQ=E(t)LQ''+RQ'+\frac{1}{C} Q=E(t)

      Which is a second order, constant coefficient, non-homogeneous differential equation.