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Applications

Solve a second-order differential equation representing simple harmonic motion.

Solve a second-order differential equation representing simple harmonic motion.

Simple Harmonic Motion

Consider a mass suspended from a spring attached to a rigid support. (This is commonly called a spring-mass system.) Gravity is pulling the mass downward and the restoring force of the spring is pulling the mass upward. As shown in , when these two forces are equal, the mass is said to be at the equilibrium position. If the mass is displaced from equilibrium, it oscillates up and down. This behavior can be modeled by a second-order constant-coefficient differential equation.

Let \(x(t)\) denote the displacement of the mass from equilibrium. Note that for spring-mass systems of this type, it is customary to adopt the convention that down is positive. Thus, a positive displacement indicates the mass is below the equilibrium point, whereas a negative displacement indicates the mass is above equilibrium. Displacement is usually given in feet in the English system or meters in the metric system.

Consider the forces acting on the mass. The force of gravity is given by \(mg\text{.}\) In the English system, mass is in slugs and the acceleration resulting from gravity is in feet per second squared. The acceleration resulting from gravity is constant, so in the English system, \(g=32\) ft/sec2. Recall that 1 slug-foot/sec2 is a pound, so the expression mg can be expressed in pounds. Metric system units are kilograms for mass and m/sec2 for gravitational acceleration. In the metric system, we have \(g=9.8\) m/sec2.

According to Hooke’s law, the restoring force of the spring is proportional to the displacement and acts in the opposite direction from the displacement, so the restoring force is given by \(\text{-}k(s+x).\) The spring constant is given in pounds per foot in the English system and in newtons per meter in the metric system.

Now, by Newton’s second law, the sum of the forces on the system (gravity plus the restoring force) is equal to mass times acceleration, so we have

\[\begin{array}{ll}m{x}^{″} & =\text{-}k(s+x)+mg \\ & =\text{-}ks-kx+mg\text{.}\end{array}\]

However, by the way we have defined our equilibrium position, \(mg=ks,\) the differential equation becomes

\[m{x}^{″}+kx=0.\]

It is convenient to rearrange this equation and introduce a new variable, called the angular frequency, \(\omega \text{.}\) Letting \(\omega =\sqrt{k\text{/}m},\) we can write the equation as

\[{x}^{″}+{\omega }^{2}x=0.\]\[x(t)={c}_{1}\text{cos}\ \omega t+{c}_{2}\text{sin}\ \omega t,\]\[x(t)=A\ \text{sin}\ (\omega t+ϕ).\]\[{c}_{1}\text{cos}\ (\omega t)+{c}_{2}\text{sin}\ (\omega t)=A\ \text{sin}\ (\omega t+ϕ).\]\[\text{sin}(\alpha +\beta )=\text{sin}\ \alpha \ \text{cos}\ \beta +\text{cos}\ \alpha \ \text{sin}\ \beta\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Damped Vibrations

With the model just described, the motion of the mass continues indefinitely. Clearly, this doesn’t happen in the real world. In the real world, there is almost always some friction in the system, which causes the oscillations to die off slowly—an effect called damping. So now let’s look at how to incorporate that damping force into our differential equation.

Physical spring-mass systems almost always have some damping as a result of friction, air resistance, or a physical damper, called a dashpot (a pneumatic cylinder; see ).

Because damping is primarily a friction force, we assume it is proportional to the velocity of the mass and acts in the opposite direction. So the damping force is given by \(\text{-}b{x}^{'}\) for some constant \(b>0.\) Again applying Newton’s second law, the differential equation becomes

\[m{x}^{″}+b{x}^{'}+kx=0.\]

Then the associated characteristic equation is

\[m{\lambda }^{2}+b\lambda +k=0.\]

Applying the quadratic formula, we have

\[\lambda =\frac{\text{-}b\pm \sqrt{{b}^{2}-4mk}}{2m}.\]

Just as in Second-Order Linear Equations we consider three cases, based on whether the characteristic equation has distinct real roots, a repeated real root, or complex conjugate roots.

Condensed — the full section is in OpenStax Calculus Volume 3.

Forced Vibrations

The last case we consider is when an external force acts on the system. In the case of the motorcycle suspension system, for example, the bumps in the road act as an external force acting on the system. Another example is a spring hanging from a support; if the support is set in motion, that motion would be considered an external force on the system. We model these forced systems with the nonhomogeneous differential equation

\[m{x}^{″}+b{x}^{'}+kx=f(t),\]

where the external force is represented by the \(f(t)\) term. As we saw in Nonhomogeneous Linear Equations, differential equations such as this have solutions of the form

\[x(t)={c}_{1}{x}_{1}(t)+{c}_{2}{x}_{2}(t)+{x}_{p}(t),\]

where \({c}_{1}{x}_{1}(t)+{c}_{2}{x}_{2}(t)\) is the general solution to the complementary equation and \({x}_{p}(t)\) is a particular solution to the nonhomogeneous equation. If the system is damped, \(\underset{t\to \infty }{\text{lim}}{c}_{1}{x}_{1}(t)+{c}_{2}{x}_{2}(t)=0.\) Since these terms do not affect the long-term behavior of the system, we call this part of the solution the transient solution. The long-term behavior of the system is determined by \({x}_{p}(t),\) so we call this part of the solution the steady-state solution.

Example

Try it.

A mass of 1 slug stretches a spring 2 ft and comes to rest at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. Find the equation of motion if an external force equal to \(f(t)=8\ \text{sin}\ (4t)\) is applied to the system beginning at time \(t=0.\) What is the transient solution? What is the steady-state solution?

Solution

We have \(mg=1(32)=2k,\) so \(k=16\) and the differential equation is

\[{x}^{″}+8{x}^{'}+16x=8\ \text{sin}\ (4t).\]

The general solution to the complementary equation is

\[{c}_{1}{e}^{-4t}+{c}_{2}t{e}^{-4t}.\]

Assuming a particular solution of the form \({x}_{p}(t)=A\ \text{cos}\ (4t)+B\ \text{sin}\ (4t)\) and using the method of undetermined coefficients, we find \({x}_{p}(t)=-\frac{1}{4}\ \text{cos}\ (4t),\) so

\[x(t)={c}_{1}{e}^{-4t}+{c}_{2}t{e}^{-4t}-\frac{1}{4}\ \text{cos}\ (4t).\]

At \(t=0,\) the mass is at rest in the equilibrium position, so \(x(0)={x}^{'}(0)=0.\) Applying these initial conditions to solve for \({c}_{1}\) and \({c}_{2},\) we get

\[x(t)=\frac{1}{4}{e}^{-4t}+t{e}^{-4t}-\frac{1}{4}\ \text{cos}\ (4t).\]

The transient solution is \(\frac{1}{4}{e}^{-4t}+t{e}^{-4t}.\) The steady-state solution is \(-\frac{1}{4}\ \text{cos}\ (4t).\)

Condensed — the full section is in OpenStax Calculus Volume 3.

The

Consider an electrical circuit containing a resistor, an inductor, and a capacitor, as shown in . Such a circuit is called an RLC series circuit. RLC circuits are used in many electronic systems, most notably as tuners in AM/FM radios. The tuning knob varies the capacitance of the capacitor, which in turn tunes the radio. Such circuits can be modeled by second-order, constant-coefficient differential equations.

Let \(I(t)\) denote the current in the RLC circuit and \(q(t)\) denote the charge on the capacitor. Furthermore, let L denote inductance in henrys (H), R denote resistance in ohms \((\text{Ω}),\) and C denote capacitance in farads (F). Last, let \(E(t)\) denote electric potential in volts (V).

Kirchhoff’s voltage rule states that the sum of the voltage drops around any closed loop must be zero. So, we need to consider the voltage drops across the inductor (denoted \({E}_{L}\)), the resistor (denoted \({E}_{R}\)), and the capacitor (denoted \({E}_{C}\)). Because the RLC circuit shown in includes a voltage source, \(E(t),\) which adds voltage to the circuit, we have \({E}_{L}+{E}_{R}+{E}_{C}=E(t).\)

We present the formulas below without further development. Those of you interested in the derivation of these formulas should consult a physics text. Using Faraday’s law and Lenz’s law, the voltage drop across an inductor can be shown to be proportional to the instantaneous rate of change of current, with proportionality constant L. Thus,

\[{E}_{L}=L\frac{dI}{dt}.\]

Next, according to Ohm’s law, the voltage drop across a resistor is proportional to the current passing through the resistor, with proportionality constant R. Therefore,

\[{E}_{R}=RI\text{.}\]

Last, the voltage drop across a capacitor is proportional to the charge, q, on the capacitor, with proportionality constant \(1\text{/}C.\) Thus,

\[{E}_{C}=\frac{1}{C}q\text{.}\]

Adding these terms together, we get

\[L\frac{dI}{dt}+RI+\frac{1}{C}q=E(t).\]\[L\frac{{d}^{2}q}{d{t}^{2}}+R\frac{dq}{dt}+\frac{1}{C}q=E(t).\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

  • Second-order constant-coefficient differential equations can be used to model spring-mass systems.
  • An examination of the forces on a spring-mass system results in a differential equation of the form
    \[m{x}^{″}+b{x}^{'}+kx=f(t),\]
    where \(m\) represents the mass, \(b\) is the coefficient of the damping force, \(k\) is the spring constant, and \(f(t)\) represents any net external forces on the system.
  • If \(b=0,\) there is no damping force acting on the system, and simple harmonic motion results. If \(b\ne 0,\) the behavior of the system depends on whether \({b}^{2}-4mk>0,\) \({b}^{2}-4mk=0,\) or \({b}^{2}-4mk<0.\)
  • If \({b}^{2}-4mk>0,\) the system is overdamped and does not exhibit oscillatory behavior.
  • If \({b}^{2}-4mk=0,\) the system is critically damped. It does not exhibit oscillatory behavior, but any slight reduction in the damping would result in oscillatory behavior.
  • If \({b}^{2}-4mk<0,\) the system is underdamped. It exhibits oscillatory behavior, but the amplitude of the oscillations decreases over time.
  • If \(f(t)\ne 0,\) the solution to the differential equation is the sum of a transient solution and a steady-state solution. The steady-state solution governs the long-term behavior of the system.
  • The charge on the capacitor in an RLC series circuit can also be modeled with a second-order constant-coefficient differential equation of the form
    \[L\frac{{d}^{2}q}{d{t}^{2}}+R\frac{dq}{dt}+\frac{1}{C}q=E(t),\]
    where L is the inductance, R is the resistance, C is the capacitance, and \(E(t)\) is the voltage source.

Key Equations

Equation of simple harmonic motion\({x}^{″}+{\omega }^{2}x=0\)
Solution for simple harmonic motion\(x(t)={c}_{1}\text{cos}\ (\omega t)+{c}_{2}\text{sin}\ (\omega t)\)
Alternative form of solution for SHM\(x(t)=A\ \text{sin}\ (\omega t+ϕ)\)
Forced harmonic motion\(m{x}^{″}+b{x}^{'}+kx=f(t)\)
Charge in a RLC series circuit\(L\frac{{d}^{2}q}{d{t}^{2}}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\)

Practice (32)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Assume an object weighing 2 lb stretches a spring 6 in. Find the equation of motion if the spring is released from the equilibrium position with an upward velocity of 16 ft/sec. What is the period of the motion?

    ჲრკპთირვ ჲრდჲგჲპა.

    We first need to find the spring constant. We have

    \[\begin{array}{lll}mg & = & ks \\ 2 & = & k(\frac{1}{2}) \\ k & = & 4.\end{array}\]

    We also know that weight W equals the product of mass m and the acceleration due to gravity g. In English units, the acceleration due to gravity is 32 ft/sec2.

    \[\begin{array}{lll}W & = & mg \\ 2 & = & m(32) \\ m & = & \frac{1}{16}\end{array}\]

    Thus, the differential equation representing this system is

    \[\frac{1}{16}{x}^{″}+4x=0.\]

    Multiplying through by 16, we get \({x}^{″}+64x=0,\) which can also be written in the form \({x}^{″}+({8}^{2})x=0.\) This equation has the general solution

    \[x(t)={c}_{1}\text{cos}\ (8t)+{c}_{2}\text{sin}\ (8t).\]

    The mass was released from the equilibrium position, so \(x(0)=0,\) and it had an initial upward velocity of 16 ft/sec, so \({x}^{'}(0)=-16.\) Applying these initial conditions to solve for \({c}_{1}\) and \({c}_{2}.\) gives

    \[x(t)=-2\ \text{sin}\ 8t\text{.}\]

    The period of this motion is \(\frac{2\pi }{8}=\frac{\pi }{4}\) sec.

  2. A 200-g mass stretches a spring 5 cm. Find the equation of motion of the mass if it is released from rest from a position 10 cm below the equilibrium position. What is the frequency of this motion?

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(t)=0.1\ \text{cos}\ (14t)\) (in meters); frequency is \(\frac{14}{2\pi }\) Hz.

  3. Express the following functions in the form \(A\ \text{sin}\ (\omega t+ϕ).\) What is the frequency of motion? The amplitude?

    1. \(x(t)=2\ \text{cos}\ (3t)+\text{sin}\ (3t)\)
    2. \(x(t)=3\ \text{cos}\ (2t)-2\ \text{sin}\ (2t)\)
    ჲრკპთირვ ჲრდჲგჲპა.
    1. We have
      \[A=\sqrt{{c}_{1}^{2}+{c}_{2}^{2}}=\sqrt{{2}^{2}+{1}^{2}}=\sqrt{5}\]
      and
      \[\text{tan}\ ϕ=\frac{{c}_{1}}{{c}_{2}}=\frac{2}{1}=2.\]
      Note that both \({c}_{1}\) and \({c}_{2}\) are positive, so \(ϕ\) is in the first quadrant. Thus,
      \[ϕ\approx 1.107\ \text{rad,}\]
      so we have
      \[x(t)=2\ \text{cos}\ (3t)+\text{sin}\ (3t)=\sqrt{5}\text{sin}\ (3t+1.107).\]
      The frequency is \(\frac{\omega }{2\pi }=\frac{3}{2\pi }\approx 0.477.\) The amplitude is \(\sqrt{5}.\)
    2. We have
      \[A=\sqrt{{c}_{1}^{2}+{c}_{2}^{2}}=\sqrt{{3}^{2}+{2}^{2}}=\sqrt{13}\]
      and
      \[\text{tan}\ ϕ=\frac{{c}_{1}}{{c}_{2}}=\frac{3}{-2}=-\frac{3}{2}.\]
      Note that \({c}_{1}\) is positive but \({c}_{2}\) is negative, so \(ϕ\) is in the fourth quadrant. Thus,
      \[ϕ\approx -0.983\ \text{rad,}\]
      so we have
      \[\begin{array}{ll}x(t) & =3\ \text{cos}\ (2t)-2\ \text{sin}\ (2t) \\ & =\sqrt{13}\text{sin}\ (2t-0.983).\end{array}\]
      The frequency is \(\frac{\omega }{2\pi }=\frac{2}{2\pi }\approx 0.318.\) The amplitude is \(\sqrt{13}.\)

  4. Express the function \(x(t)=\text{cos}\ (4t)+4\ \text{sin}\ (4t)\) in the form \(A\ \text{sin}\ (\omega t+ϕ).\) What is the frequency of motion? The amplitude?

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(t)=\sqrt{17}\text{sin}\ (4t+0.245),\) \(\text{frequency}=\frac{4}{2\pi }\approx 0.637,\) \(A=\sqrt{17}\)

  5. A 16-lb mass is attached to a 10-ft spring. When the mass comes to rest in the equilibrium position, the spring measures 15 ft 4 in. The system is immersed in a medium that imparts a damping force equal to \(\frac{5}{2}\) times the instantaneous velocity of the mass. Find the equation of motion if the mass is pushed upward from the equilibrium position with an initial upward velocity of 5 ft/sec. What is the position of the mass after 10 sec? Its velocity?

    ჲრკპთირვ ჲრდჲგჲპა.

    The mass stretches the spring 5 ft 4 in., or \(\frac{16}{3}\) ft. Thus, \(16=(\frac{16}{3})k,\) so \(k=3.\) We also have \(m=\frac{16}{32}=\frac{1}{2},\) so the differential equation is

    \[\frac{1}{2}{x}^{″}+\frac{5}{2}{x}^{'}+3x=0.\]

    Multiplying through by 2 gives \({x}^{″}+5{x}^{'}+6x=0,\) which has the general solution

    \[x(t)={c}_{1}{e}^{-2t}+{c}_{2}{e}^{-3t}.\]

    Applying the initial conditions, \(x(0)=0\) and \({x}^{'}(0)=-5,\) we get

    \[x(t)=-5{e}^{-2t}+5{e}^{-3t}.\]

    After 10 sec the mass is at position

    \[x(10)=-5{e}^{-20}+5{e}^{-30}\approx -1.0305\ \times \ {10}^{-8}\approx 0,\]

    so it is, effectively, at the equilibrium position. We have \({x}^{'}(t)=10{e}^{-2t}-15{e}^{-3t},\) so after 10 sec the mass is moving at a velocity of

    \[{x}^{'}(10)=10{e}^{-20}-15{e}^{-30}\approx 2.061\ \times \ {10}^{-8}\approx 0.\]

    After only 10 sec, the mass is barely moving.

  6. A 2-kg mass is attached to a spring with spring constant 24 N/m. The system is then immersed in a medium imparting a damping force equal to 16 times the instantaneous velocity of the mass. Find the equation of motion if it is released from rest at a point 40 cm below equilibrium.

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(t)=0.6{e}^{-2t}-0.2{e}^{-6t}\)

  7. A 1-kg mass stretches a spring 20 cm. The system is attached to a dashpot that imparts a damping force equal to 14 times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from equilibrium with an upward velocity of 3 m/sec.

    ჲრკპთირვ ჲრდჲგჲპა.

    We have \(mg=1(9.8)=0.2k,\) so \(k=49.\) Then, the differential equation is

    \[{x}^{″}+14{x}^{'}+49x=0,\]

    which has general solution

    \[x(t)={c}_{1}{e}^{-7t}+{c}_{2}t{e}^{-7t}.\]

    Applying the initial conditions \(x(0)=0\) and \({x}^{'}(0)=-3\) gives

    \[x(t)=-3t{e}^{-7t}.\]

  8. A 1-lb weight stretches a spring 6 in., and the system is attached to a dashpot that imparts a damping force equal to half the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 6 in. below equilibrium.

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(t)=\frac{1}{2}{e}^{-8t}+4t{e}^{-8t}\)

  9. A 16-lb weight stretches a spring 3.2 ft. Assume the damping force on the system is equal to the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 9 in. below equilibrium.

    ჲრკპთირვ ჲრდჲგჲპა.

    We have \(k=\frac{16}{3.2}=5\) and \(m=\frac{16}{32}=\frac{1}{2},\) so the differential equation is

    \[\frac{1}{2}{x}^{″}+{x}^{'}+5x=0,\ \text{or}\ {x}^{″}+2{x}^{'}+10x=0.\]

    This equation has the general solution

    \[x(t)={e}^{\text{-}t}({c}_{1}\text{cos}\ (3t)+{c}_{2}\text{sin}\ (3t)).\]

    Applying the initial conditions, \(x(0)=\frac{3}{4}\) and \({x}^{'}(0)=0,\) we get

    \[x(t)={e}^{\text{-}t}(\frac{3}{4}\ \text{cos}\ (3t)+\frac{1}{4}\ \text{sin}\ (3t)).\]

  10. A 1-kg mass stretches a spring 49 cm. The system is immersed in a medium that imparts a damping force equal to four times the instantaneous velocity of the mass. Find the equation of motion if the mass is released from rest at a point 24 cm above equilibrium.

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(t)=-0.24{e}^{-2t}\text{cos}\ (4t)-0.12{e}^{-2t}\text{sin}\ (4t)\)

  11. A mass of 1 slug stretches a spring 2 ft and comes to rest at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. Find the equation of motion if an external force equal to \(f(t)=8\ \text{sin}\ (4t)\) is applied to the system beginning at time \(t=0.\) What is the transient solution? What is the steady-state solution?

    ჲრკპთირვ ჲრდჲგჲპა.

    We have \(mg=1(32)=2k,\) so \(k=16\) and the differential equation is

    \[{x}^{″}+8{x}^{'}+16x=8\ \text{sin}\ (4t).\]

    The general solution to the complementary equation is

    \[{c}_{1}{e}^{-4t}+{c}_{2}t{e}^{-4t}.\]

    Assuming a particular solution of the form \({x}_{p}(t)=A\ \text{cos}\ (4t)+B\ \text{sin}\ (4t)\) and using the method of undetermined coefficients, we find \({x}_{p}(t)=-\frac{1}{4}\ \text{cos}\ (4t),\) so

    \[x(t)={c}_{1}{e}^{-4t}+{c}_{2}t{e}^{-4t}-\frac{1}{4}\ \text{cos}\ (4t).\]

    At \(t=0,\) the mass is at rest in the equilibrium position, so \(x(0)={x}^{'}(0)=0.\) Applying these initial conditions to solve for \({c}_{1}\) and \({c}_{2},\) we get

    \[x(t)=\frac{1}{4}{e}^{-4t}+t{e}^{-4t}-\frac{1}{4}\ \text{cos}\ (4t).\]

    The transient solution is \(\frac{1}{4}{e}^{-4t}+t{e}^{-4t}.\) The steady-state solution is \(-\frac{1}{4}\ \text{cos}\ (4t).\)

  12. A mass of 2 kg is attached to a spring with constant 32 N/m and comes to rest in the equilibrium position. Beginning at time \(t=0,\) an external force equal to\(f(t)=68{e}^{-2t}\text{cos}\ (4t)\) is applied to the system. Find the equation of motion if there is no damping. What is the transient solution? What is the steady-state solution?

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(t)=-\frac{1}{2}\ \text{cos}\ (4t)+\frac{9}{4}\ \text{sin}\ (4t)+\frac{1}{2}{e}^{-2t}\text{cos}\ (4t)-2{e}^{-2t}\text{sin}\ (4t)\)
    \(\text{Transient solution:}\ \frac{1}{2}{e}^{-2t}\text{cos}\ (4t)-2{e}^{-2t}\text{sin}\ (4t)\)
    \(\text{Steady-state solution:}\ -\frac{1}{2}\ \text{cos}\ (4t)+\frac{9}{4}\ \text{sin}\ (4t)\)

  13. Find the charge on the capacitor in an RLC series circuit where \(L=5\text{/}3\) H, \(R=10\text{Ω},\) \(C=1\text{/}30\) F, and \(E(t)=300\) V. Assume the initial charge on the capacitor is 0 C and the initial current is 9 A. What happens to the charge on the capacitor over time?

    ჲრკპთირვ ჲრდჲგჲპა.

    We have

    \[\begin{array}{lll} \\ \\ L\frac{{d}^{2}q}{d{t}^{2}}+R\frac{dq}{dt}+\frac{1}{C}q & = & E(t) \\ \frac{5}{3}\ \frac{{d}^{2}q}{d{t}^{2}}+10\frac{dq}{dt}+30q & = & 300 \\ \frac{{d}^{2}q}{d{t}^{2}}+6\frac{dq}{dt}+18q & = & 180.\end{array}\]

    The general solution to the complementary equation is

    \[{e}^{-3t}({c}_{1}\text{cos}\ (3t)+{c}_{2}\text{sin}\ (3t)).\]

    Assume a particular solution of the form \({q}_{p}=A,\) where \(A\) is a constant. Using the method of undetermined coefficients, we find \(A=10.\) So,

    \[q(t)={e}^{-3t}({c}_{1}\text{cos}\ (3t)+{c}_{2}\text{sin}\ (3t))+10.\]

    Applying the initial conditions \(q(0)=0\) and \(i(0)=((dq)\text{/}(dt))(0)=9,\) we find \({c}_{1}=-10\) and \({c}_{2}=-7.\) So the charge on the capacitor is

    \[q(t)=-10{e}^{-3t}\text{cos}\ (3t)-7{e}^{-3t}\text{sin}\ (3t)+10.\]

    Looking closely at this function, we see the first two terms will decay over time (as a result of the negative exponent in the exponential function). Therefore, the capacitor eventually approaches a steady-state charge of 10 C.

  14. Find the charge on the capacitor in an RLC series circuit where \(L=1\text{/}5\) H, \(R=2\text{/}5\text{Ω},\) \(C=1\text{/}2\) F, and \(E(t)=50\) V. Assume the initial charge on the capacitor is 0 C and the initial current is 4 A.

    ჲრკპთირვ ჲრდჲგჲპა.

    \(q(t)=-25{e}^{\text{-}t}\text{cos}\ (3t)-7{e}^{\text{-}t}\text{sin}\ (3t)+25\)

  15. A mass weighing 4 lb stretches a spring 8 in. Find the equation of motion if the spring is released from the equilibrium position with a downward velocity of 12 ft/sec. What is the period and frequency of the motion?

  16. A mass weighing 2 lb stretches a spring 2 ft. Find the equation of motion if the spring is released from 2 in. below the equilibrium position with an upward velocity of 8 ft/sec. What is the period and frequency of the motion?

    ჲრკპთირვ ჲრდჲგჲპა.

    \({x}^{″}+16x=0,\) \(x(t)=\frac{1}{6}\ \text{cos}\ (4t)-2\ \text{sin}\ (4t)\text{,}\) period \(=\frac{\pi }{2}\ \text{sec},\) frequency \(=\frac{2}{\pi }\ \text{Hz}\)

  17. A 100-g mass stretches a spring 0.1 m. Find the equation of motion of the mass if it is released from rest from a position 20 cm below the equilibrium position. What is the frequency of this motion?

  18. A 400-g mass stretches a spring 5 cm. Find the equation of motion of the mass if it is released from rest from a position 15 cm below the equilibrium position. What is the frequency of this motion?

    ჲრკპთირვ ჲრდჲგჲპა.

    \({x}^{″}+196x=0,\) \(x(t)=0.15\ \text{cos}\ (14t)\text{,}\) period \(=\frac{\pi }{7}\ \text{sec},\) frequency \(=\frac{7}{\pi }\ \text{Hz}\)

  19. A block has a mass of 9 kg and is attached to a vertical spring with a spring constant of 0.25 N/m. The block is stretched 0.75 m below its equilibrium position and released.

    1. Find the position function \(x(t)\) of the block.
    2. Find the period and frequency of the vibration.
    3. Sketch a graph of \(x(t).\)
    4. At what time does the block first pass through the equilibrium position?

  20. A block has a mass of 5 kg and is attached to a vertical spring with a spring constant of 20 N/m. The block is released from the equilibrium position with a downward velocity of 10 m/sec.

    1. Find the position function \(x(t)\) of the block.
    2. Find the period and frequency of the vibration.
    3. Sketch a graph of \(x(t).\)
    4. At what time does the block first pass through the equilibrium position?
    ჲრკპთირვ ჲრდჲგჲპა.

    a. \(x(t)=5\ \text{sin}\ (2t)\)
    b. period \(=\pi \ \text{sec},\) frequency \(=\frac{1}{\pi }\ \text{Hz}\)
    c.

    d. \(t=\frac{\pi }{2}\ \text{sec}\)

  21. A 1-kg mass is attached to a vertical spring with a spring constant of 21 N/m. The resistance in the spring-mass system is equal to 10 times the instantaneous velocity of the mass.

    1. Find the equation of motion if the mass is released from a position 2 m below its equilibrium position with a downward velocity of 2 m/sec.
    2. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.

  22. An 800-lb weight (25 slugs) is attached to a vertical spring with a spring constant of 226 lb/ft. The system is immersed in a medium that imparts a damping force equal to 10 times the instantaneous velocity of the mass.

    1. Find the equation of motion if it is released from a position 20 ft below its equilibrium position with a downward velocity of 41 ft/sec.
    2. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.
    ჲრკპთირვ ჲრდჲგჲპა.

    a. \(x(t)={e}^{\text{-}t\text{/}5}(20\ \text{cos}\ (3t)+15\ \text{sin}\ (3t))\)
    b. underdamped

  23. A 9-kg mass is attached to a vertical spring with a spring constant of 16 N/m. The system is immersed in a medium that imparts a damping force equal to 24 times the instantaneous velocity of the mass.

    1. Find the equation of motion if it is released from its equilibrium position with an upward velocity of 4 m/sec.
    2. Graph the solution and determine whether the motion is overdamped, critically damped, or underdamped.

  24. A 1-kg mass stretches a spring 61.25 cm. The resistance in the spring-mass system is equal to eight times the instantaneous velocity of the mass.

    1. Find the equation of motion if the mass is released from a position 5 m below its equilibrium position with an upward velocity of 10 m/sec.
    2. Determine whether the motion is overdamped, critically damped, or underdamped.
    ჲრკპთირვ ჲრდჲგჲპა.

    a. \(x(t)={e}^{-4t}(-5\cos 11.866t-0.834\sin 11.866t)\)
    b. underdamped

  25. A 32-lb weight (1 slug) stretches a vertical spring 128 in. The resistance in the spring-mass system is equal to four times the instantaneous velocity of the mass.

    1. Find the equation of motion if it is released from its equilibrium position with a downward velocity of 12 ft/sec.
    2. Determine whether the motion is overdamped, critically damped, or underdamped.

  26. A 64-lb weight is attached to a vertical spring with a spring constant of 4.625 lb/ft. The resistance in the spring-mass system is equal to the instantaneous velocity. The weight is set in motion from a position 1 ft below its equilibrium position with an upward velocity of 2 ft/sec. Is the mass above or below the equilibrium position at the end of \(\pi\) sec? By what distance?

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(\pi )=\frac{7{e}^{\text{-}\pi \text{/}4}}{6}\) ft below

  27. A mass that weighs 8 lb stretches a spring 6 inches. The system is acted on by an external force of \(8\ \text{sin}\ 8t\) lb. If the mass is pulled down 3 inches and then released, determine the position of the mass at any time.

  28. A mass that weighs 6 lb stretches a spring 3 in. The system is acted on by an external force of \(8\ \text{sin}\ (4t)\) lb. If the mass is pulled down 1 inch and then released, determine the position of the mass at any time.

    ჲრკპთირვ ჲრდჲგჲპა.

    \(x(t)=\frac{1}{2}\ \text{cos}\ (8\sqrt{2}t)-\frac{4\sqrt{2}}{21}\text{sin}\left(8\sqrt{2}t\right)+\frac{8}{21}\text{sin}\left(8\sqrt{2}t\right)\)

  29. Find the charge on the capacitor in an RLC series circuit where \(L=40\) H, \(R=30\text{Ω},\) \(C=1\text{/}200\) F, and \(E(t)=200\) V. Assume the initial charge on the capacitor is 7 C and the initial current is 0 A.

  30. Find the charge on the capacitor in an RLC series circuit where \(L=2\) H, \(R=24\text{Ω},\) \(C=0.005\) F, and \(E(t)=12\ \text{sin}\ 10t\) V. Assume the initial charge on the capacitor is 0.001 C and the initial current is 0 A.

    ჲრკპთირვ ჲრდჲგჲპა.

    \(q(t)={e}^{-6t}(0.051\ \text{cos}\ (8t)+0.03825\ \text{sin}\ (8t))-\frac{1}{20}\ \text{cos}\ (10t)\)

  31. A series circuit consists of a device where \(L=1\) H, \(R=20\text{Ω},\) \(C=0.002\) F, and \(E(t)=12\) V. If the initial charge and current are both zero, find the charge and current at time t.

  32. A series circuit consists of a device where \(L=\frac{1}{2}\) H, \(R=10\text{Ω},\) \(C=\frac{1}{50}\) F, and \(E(t)=250\) V. If the initial charge on the capacitor is 0 C and the initial current is 18 A, find the charge and current at time t.

    ჲრკპთირვ ჲრდჲგჲპა.

    \(q(t)={e}^{-10t}(-32t-5)+5,\ \ I(t)=2{e}^{-10t}(160t+9)\)

Symbols used here

\sqrt{x},\ \sqrt[n]{x}
square root, n-th root
The non-negative number whose square (n-th power) is x.
\infty
infinity
Not a number: "grows without bound" in limits and intervals.
\lambda
lambda (eigenvalue)
The factor by which an eigenvector is stretched: Av = λv.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
\pm
plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.
\neq
not equal
The two sides are different.
\sup,\ \inf
supremum, infimum
Least upper bound, greatest lower bound.
e
Euler's number
2.71828…, the base whose exponential is its own derivative.
\lim_{x \to a} f(x)
limit
The value f(x) approaches as x approaches a.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
y',\ y''
first and second derivative of y
Prime notation for derivatives with respect to x (or t).
C,\ C_1,\ C_2
arbitrary constants
Constants of integration fixed by initial conditions.
\frac{\partial u}{\partial t},\ \nabla^2 u
partial derivative in time, Laplacian
Rate of change in time; sum of second partials (the diffusion operator).

How to: Applications

  1. Solve a second-order differential equation representing simple harmonic motion.
  2. Solve a second-order differential equation representing damped simple harmonic motion.
  3. Solve a second-order differential equation representing forced simple harmonic motion.
  4. Solve a second-order differential equation representing charge and current in an RLC series circuit.
  5. We have
  6. We have
  7. Set up the differential equation that models the behavior of the motorcycle suspension system.
  8. We are interested in what happens when the motorcycle lands after taking a jump. Let time

Questions people ask

What is a differential equation?

An equation whose unknown is a function, relating it to its own derivatives. "The rate of growth is proportional to the population" is y′ = ky, and solving it means finding y as a function of time.

Why does the solution have arbitrary constants?

Integrating loses information: many functions share the same derivative. An n-th order equation has n constants, fixed by n initial or boundary conditions.

ჲოთრაი ჟამ.

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

მეტი Differential Equations

Separable equationsFirst-order linear equationsSecond-order, constant coefficientsNonhomogeneous equationsModelling with differential equations