Consider Again a Mass M on a Spring With Constant K and Damping Constant

17.three: Applications of 2d-Order Differential Equations

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Learning Objectives

• Solve a second-order differential equation representing uncomplicated harmonic movement.
• Solve a second-order differential equation representing damped elementary harmonic motility.
• Solve a second-order differential equation representing forced simple harmonic motility.
• Solve a second-order differential equation representing charge and current in an RLC series circuit.

We saw in the chapter introduction that 2nd-order linear differential equations are used to model many situations in physics and technology. In this section, we look at how this works for systems of an object with mass attached to a vertical spring and an electric circuit containing a resistor, an inductor, and a capacitor connected in series. Models such as these can be used to approximate other more complicated situations; for example, bonds betwixt atoms or molecules are often modeled as springs that vibrate, as described by these same differential equations.

Simple Harmonic Movement

Consider a mass suspended from a spring attached to a rigid support. (This is commonly called a bound-mass system.) Gravity is pulling the mass downward and the restoring forcefulness of the spring is pulling the mass upward. Equally shown in Figure \(\PageIndex{i}\), 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 downwardly. This behavior can be modeled past a second-society constant-coefficient differential equation.

Figure \(\PageIndex{1}\): A spring in its natural position (a), at equilibrium with a mass thousand attached (b), and in oscillatory movement (c).

Let \(x(t)\) announce the deportation of the mass from equilibrium. Note that for spring-mass systems of this type, information technology is customary to prefer the convention that downwardly is positive. Thus, a positive displacement indicates the mass is below the equilibrium point, whereas a negative displacement indicates the mass is to a higher place equilibrium. Displacement is usually given in anxiety in the English organisation or meters in the metric system.

Consider the forces acting on the mass. The force of gravity is given by mg.mg. In the English language arrangement, mass is in slugs and the acceleration resulting from gravity is in feet per 2d squared. The acceleration resulting from gravity is constant, then in the English organisation, \(chiliad=32\, ft/sec^two\). Call up that one slug-foot/sec² is a pound, so the expression mg tin be expressed in pounds. Metric system units are kilograms for mass and m/sec² for gravitational acceleration. In the metric organization, nosotros have \(g=9.8\) 1000/secⁱⁱ.

According to Hooke's police force, the restoring forcefulness of the spring is proportional to the displacement and acts in the reverse direction from the deportation, and then the restoring forcefulness is given by \(−k(s+x).\) The spring constant is given in pounds per human foot in the English language organization and in newtons per meter in the metric system.

At present, 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

\[\begin{align*}mx″ &=−thousand(s+ten)+mg \\[4pt] &=−ks−kx+mg. \end{align*}\]

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

\[mx″+kx=0. \nonumber\]

Information technology is convenient to rearrange this equation and introduce a new variable, called the angular frequency, \(ω\). Letting \(ω=\sqrt{chiliad/thou}\), nosotros can write the equation equally

\[x''+ω^2x=0. \nonumber\]

This differential equation has the full general solution

\[x(t)=c_1 \cos ωt+c_2 \sin ωt, \label{GeneralSol}\]

which gives the position of the mass at whatever bespeak in time. The motion of the mass is called simple harmonic motion. The catamenia of this movement (the time it takes to complete one oscillation) is \(T=\dfrac{2π}{ω}\) and the frequency is \(f=\dfrac{ane}{T}=\dfrac{ω}{2π}\) (Figure \(\PageIndex{2}\)).

Figure \(\PageIndex{two}\): A graph of vertical displacement versus time for simple harmonic move.

Example \(\PageIndex{1}\): Unproblematic Harmonic Motion

Presume an object weighing ii lb stretches a spring vi in. Find the equation of motility if the leap is released from the equilibrium position with an upward velocity of 16 ft/sec. What is the period of the movement?

Solution

We kickoff need to notice the spring constant. Nosotros take

\[\brainstorm{align*}mg &=ks\\[4pt] 2 &=1000 \left(\dfrac{i}{2}\right)\\[4pt] k &=four. \end{align*}\]

We also know that weight \(West\) equals the production of mass \(grand\) and the acceleration due to gravity \(m\). In English language units, the dispatch due to gravity is 32 ft/sec^two.

\[\brainstorm{align*}Westward &=mg\\[4pt] 2 &=m(32)\\[4pt] m &=\dfrac{ane}{16}\end{align*}\]

Thus, the differential equation representing this arrangement is

\[\dfrac{1}{16}x″+4x=0. \nonumber\]

Multiplying through past 16, we get \(x''+64x=0,\) which can likewise be written in the form \(ten''+(viii^2)x=0.\) This equation has the general solution

\[x(t)=c_1 \cos (8t)+c_2 \sin (8t). \nonumber\]

The mass was released from the equilibrium position, then \(10(0)=0\), and information technology had an initial upward velocity of 16 ft/sec, so \(ten′(0)=−16\). Applying these initial conditions to solve for \(c_1\) and \(c_2\). gives

\[x(t)=−two \sin 8t. \nonumber\]

The flow of this move is \(\dfrac{2π}{8}=\dfrac{π}{4}\) sec.

Do \(\PageIndex{1}\)

A 200-chiliad mass stretches a bound 5 cm. Find the equation of motion of the mass if it is released from rest from a position x cm below the equilibrium position. What is the frequency of this motility?

Hint

First find the spring constant.

Answer

\(10(t)=0.i \cos (14t)\) (in meters); frequency is \(\dfrac{xiv}{2π}\) Hz.

Writing the general solution in the form \(x(t)=c_1 \cos (ωt)+c_2 \sin(ωt)\) (Equation \ref{GeneralSol}) has some advantages. Information technology is easy to see the link between the differential equation and the solution, and the menstruum and frequency of motion are evident. This form of the function tells u.s.a. very little about the amplitude of the motion, however. In some situations, we may prefer to write the solution in the form

\[x(t)=A \sin (ωt+ϕ). \nonumber\]

Although the link to the differential equation is not as explicit in this case, the catamenia and frequency of motility are even so evident. Furthermore, the amplitude of the motion, \(A,\) is obvious in this form of the office. The abiding \(ϕ\) is called a phase shift and has the upshot of shifting the graph of the function to the left or right.

To convert the solution to this form, we want to observe the values of \(A\) and \(ϕ\) such that

\[c_1 \cos (ωt)+c_2 \sin (ωt)=A \sin (ωt+ϕ). \nonumber\]

We first apply the trigonometric identity

\[\sin (α+β)= \sin α \cos β+ \cos α \sin β \nonumber\]

to get

\[\begin{marshal*} c_1 \cos (ωt)+c_2 \sin (ωt) &= A( \sin (ωt) \cos ϕ+ \cos (ωt) \sin ϕ) \\[4pt] &= A \sin ϕ( \cos (ωt))+A \cos ϕ( \sin (ωt)). \end{align*}\]

Thus,

\[c1=A \sin ϕ \text{ and } c_2=A \cos ϕ. \nonumber\]

If nosotros square both of these equations and add them together, we go

\[\begin{align*}c_1^2+c_2^two &=A^2 \sin _2 ϕ+A^ii \cos _2 ϕ \\[4pt] &=A^ii( \sin ^2 ϕ+ \cos ^2 ϕ) \\[4pt] &=A^ii. \end{align*}\]

Thus,

\[A=\sqrt{c_1^2+c_2^2}. \nonumber\]

Now, to discover \(ϕ\), become back to the equations for \(c_1\) and \(c_2\), but this fourth dimension, split up the get-go equation by the second equation to get

\[\begin{align*} \dfrac{c_1}{c_2} &=\dfrac{A \sin ϕ}{A \cos ϕ} \\[4pt] &= \tan ϕ. \end{marshal*} \]

Then,

\[\tan ϕ= \dfrac{c_1}{c_2}. \nonumber\]

We summarize this finding in the post-obit theorem.

Solution TO THE EQUATION FOR SIMPLE HARMONIC Movement

The function \(x(t)=c_1 \cos (ωt)+c_2 \sin (ωt)\) tin exist written in the grade \(10(t)=A \sin (ωt+ϕ)\), where \(A=\sqrt{c_1^2+c_2^2}\) and \( \tan ϕ = \dfrac{c_1}{c_2}\).

Note that when using the formula \( \tan ϕ=\dfrac{c_1}{c_2}\) to discover \(ϕ\), we must accept care to ensure \(ϕ\) is in the right quadrant (Figure \(\PageIndex{3}\)).

Figure \(\PageIndex{3}\): A graph of vertical displacement versus time for simple harmonic movement with a phase alter.

Instance \(\PageIndex{2}\): Expressing the Solution with a Phase Shift

Express the following functions in the course \(A \sin (ωt+ϕ) \). What is the frequency of motion? The aamplitude?

1. \(10(t)=2 \cos (3t)+ \sin (3t)\)
2. \(10(t)=3 \cos (2t)−2 \sin (2t)\)

Solution

We accept

\[A=\sqrt{c_1^2+c_2^2}=\sqrt{2^two+1^2}=\sqrt{5} \nonumber\]

and

\[ \tan ϕ = \dfrac{c_1}{c_2}=\dfrac{2}{ane}=2. \nonumber\]

Annotation that both \(c_1\) and \(c_2\) are positive, so \(ϕ\) is in the first quadrant. Thus,

\[ϕ≈ane.107 \; \text{rad}, \nonumber\]

so nosotros have

\[ x(t) = 2 \cos (3t)+ \sin (3t) =five \sin (3t+1.107). \nonumber\]

The frequency is \(\dfrac{ω}{2π}=\dfrac{3}{2π}≈0.477.\) The amplitude is \(\sqrt{5}\).

1. We accept

   \[A=\sqrt{c_1^2+c_2^2}=\sqrt{three^2+2^2}=\sqrt{13} \nonumber\]

   and

   \[ \tan ϕ = \dfrac{c_1}{c_2}= \dfrac{3}{−two}=−\dfrac{3}{two}. \nonumber\]

   Notation that \(c_1\) is positive but \(c_2\) is negative, so \(ϕ\) is in the quaternary quadrant. Thus,

   \[ϕ≈−0.983 \; \text{rad}, \nonumber\]

   so we have

   \[\begin{align*} x(t) &=iii \cos (2t) −ii \sin (2t) \\ &= \sqrt{13} \sin (2t−0.983). \stop{align*}\]

   The frequency is \(\dfrac{ω}{2π}=\dfrac{two}{2π}≈0.318.\) The amplitude is \(\sqrt{thirteen}\).

Exercise \(\PageIndex{2}\)

Limited the function \(x(t)= \cos (4t) + 4 \sin (4t)\) in the form \(A \sin (ωt+ϕ) \). What is the frequency of movement? The aamplitude?

Hint

Use the process from the Example \(\PageIndex{ii}\).

Answer

\(10(t)= \sqrt{17} \sin (4t+0.245), \text{frequency} =\dfrac{iv}{2π}≈0.637, A=\sqrt{17}\)

Damped Vibrations

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

Physical leap-mass systems well-nigh always have some damping equally a result of friction, air resistance, or a physical damper, chosen a dashpot (a pneumatic cylinder; Effigy \(\PageIndex{4}\)).

Effigy \(\PageIndex{4}\): A dashpot is a pneumatic cylinder that dampens the motion of an aquiver system.

Because damping is primarily a friction force, nosotros assume it is proportional to the velocity of the mass and acts in the opposite management. And then the damping force is given by \(−bx′\) for some abiding \(b>0\). Over again applying Newton'due south second law, the differential equation becomes

\[mx″+bx′+kx=0. \nonumber\]

And then the associated characteristic equation is

\[mλ^ii+bλ+grand=0. \nonumber\]

Applying the quadratic formula, we have

\[λ=\dfrac{−b±\sqrt{b^two−4mk}}{2m}. \nonumber\]

Merely every bit in Second-Order Linear Equations we consider iii cases, based on whether the feature equation has singled-out real roots, a repeated real root, or circuitous conjugate roots.

Instance i: Overdamped Vibrations

When \(b^2>4mk\), we say the arrangement is overdamped . The general solution has the form

\[x(t)=c_1e^{λ_1t}+c_2e^{λ_2t}, \nonumber\]

where both \(λ_1\) and \(λ_2\) are less than zilch. Because the exponents are negative, the displacement decays to zero over fourth dimension, normally quite chop-chop. Overdamped systems practice non oscillate (no more than than i change of direction), merely only move dorsum toward the equilibrium position. Figure \(\PageIndex{5}\) shows what typical critically damped beliefs looks like.

Figure \(\PageIndex{5}\): Behavior of an overdamped spring-mass arrangement, with no change in direction (a) and just one change in direction (b).

Instance \(\PageIndex{3}\): Overdamped Leap-Mass System

A 16-lb mass is attached to a 10-ft spring. When the mass comes to balance in the equilibrium position, the spring measures 15 ft 4 in. The system is immersed in a medium that imparts a damping strength equal to 5252 times the instantaneous velocity of the mass. Find the equation of motion if the mass is pushed upwards 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?

Solution

The mass stretches the spring 5 ft 4 in., or \(\dfrac{16}{iii}\) ft. Thus, \(xvi=\left(\dfrac{16}{iii}\correct)thousand,\) so \(k=3.\) We as well have \(g=\dfrac{16}{32}=\dfrac{i}{ii}\), then the differential equation is

\[\dfrac{5}{two}x′+3x=0. \nonumber\]

Multiplying through by two gives \(10″+5x′+6x=0\), which has the full general solution

\[x(t)=c_1e^{−2t}+c_2e^{−3t}. \nonumber\]

Applying the initial weather, \(10(0)=0\) and \(x′(0)=−5\), nosotros get

\[ten(t)=−5e^{−2t}+5e^{−3t}. \nonumber\]

Later on ten sec the mass is at position

\[ten(x)=−5e^{−20}+5e^{−30}≈−1.0305×10^{−8}≈0, \nonumber\]

so it is, effectively, at the equilibrium position. We have \(x′(t)=10e^{−2t}−15e^{−3t}\), so afterward 10 sec the mass is moving at a velocity of

\[x′(10)=10e^{−20}−15e^{−thirty}≈two.061×10^{−8}≈0. \nonumber\]

After simply 10 sec, the mass is barely moving.

Exercise \(\PageIndex{iii}\)

A 2-kg mass is attached to a spring with spring constant 24 North/1000. 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 balance at a point 40 cm beneath equilibrium.

Hint

Follow the procedure from the previous case.

Reply

\(x(t)=0.6e^{−2t}−0.2e^{−6t}\)

Case 2: Critically Damped Vibrations

When \(b^two=4mk\), we say the system is critically damped. The general solution has the form

\[ten(t)=c_1e^{λ_1t}+c_2te^{λ_1t}, \nonumber\]

where \(λ_1\) is less than zero. The motion of a critically damped organization is very similar to that of an overdamped organization. Information technology does not oscillate. However, with a critically damped organisation, if the damping is reduced even a little, oscillatory behavior results. From a applied perspective, concrete systems are almost always either overdamped or underdamped (example three, which nosotros consider side by side). Information technology is impossible to fine-tune the characteristics of a physical system and so that \(b^2\) and \(4mk\) are exactly equal. Figure \(\PageIndex{6}\) shows what typical critically damped behavior looks like.

Figure \(\PageIndex{vi}\): Behavior of a critically damped jump-mass organization. The arrangement graphed in office (a) has more damping than the system graphed in part (b).

Example \(\PageIndex{four}\): Critically Damped Spring-Mass System

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

​​​​​​Solution

We have \(mg=1(9.viii)=0.2k\), then \(k=49.\) And then, the differential equation is

\[ten″+14x′+49x=0, \nonumber\]

which has full general solution

\[ten(t)=c_1e^{−7t}+c_2te^{−7t}. \nonumber\]

Applying the initial weather condition \(ten(0)=0\) and \(ten′(0)=−3\) gives

\[x(t)=−3te^{−7t}. \nonumber\]

Practise \(\PageIndex{4}\)

A 1-lb weight stretches a spring vi in., and the arrangement is attached to a dashpot that imparts a damping force equal to half the instantaneous velocity of the mass. Find the equation of movement if the mass is released from residual at a indicate 6 in. below equilibrium.

Hint

Get-go find the leap abiding.

Answer

\(10(t)=\dfrac{1}{2}east^{−8t}+4te^{−8t} \)

Example 3: Undamped Vibrations

When \(b^ii<4mk\), we say the organisation is underdamped. The general solution has the form

\[x(t)=east^{αt}(c_1 \cos (βt) + c_2 \sin (βt)), \nonumber\]

where \(α\) is less than zero. Underdamped systems practise oscillate because of the sine and cosine terms in the solution. However, the exponential term dominates eventually, so the amplitude of the oscillations decreases over time. Figure \(\PageIndex{7}\) shows what typical underdamped behavior looks like.

Effigy \(\PageIndex{seven}\): Behavior of an underdamped spring-mass system.

Note that for all damped systems, \( \lim \limits_{t \to \infty} x(t)=0\). The organization always approaches the equilibrium position over time.

Instance \(\PageIndex{5}\): Underdamped Spring-Mass System

A 16-lb weight stretches a leap iii.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.

Solution

Nosotros have \(grand=\dfrac{16}{three.2}=v\) and \(one thousand=\dfrac{sixteen}{32}=\dfrac{ane}{2},\) so the differential equation is

\[\dfrac{ane}{2} x″+x′+5x=0, \; \text{or} \; x″+2x′+10x=0. \nonumber\]

This equation has the general solution

\[x(t)=eastward^{−t} ( c_1 \cos (3t)+c_2 \sin (3t) ) . \nonumber\]

Applying the initial conditions, \(x(0)=\dfrac{3}{iv}\) and \(x′(0)=0,\) we become

\[10(t)=due east^{−t} \bigg( \dfrac{3}{iv} \cos (3t)+ \dfrac{one}{4} \sin (3t) \bigg) . \nonumber\]

Exercise \(\PageIndex{5}\)

A i-kg mass stretches a bound 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 movement if the mass is released from rest at a point 24 cm above equilibrium.

Hint

First find the leap constant.

Answer

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

Case \(\PageIndex{half dozen}\): Chapter Opener: Modeling a Motorcycle Suspension Organization

For motocross riders, the break systems on their motorcycles are very important. The off-route courses on which they ride often include jumps, and losing command of the motorbike when they land could toll them the race.

Figure \(\PageIndex{8}\): (credit: modification of work past nSeika, Flickr)

This suspension system can exist modeled as a damped spring-mass arrangement. We define our frame of reference with respect to the frame of the motorcycle. Assume the end of the shock absorber attached to the motorcycle frame is fixed. Then, the "mass" in our spring-mass system is the motorcycle cycle. Nosotros measure the position of the wheel with respect to the motorbike frame. This may seem counterintuitive, since, in many cases, it is really the motorcycle frame that moves, but this frame of reference preserves the development of the differential equation that was done earlier. Equally with before development, nosotros define the downward management to be positive.

When the motorcycle is lifted past its frame, the wheel hangs freely and the spring is uncompressed. This is the spring'due south natural position. When the motorcycle is placed on the ground and the rider mounts the motorcycle, the spring compresses and the system is in the equilibrium position (Figure \(\PageIndex{nine}\)).

Figure \(\PageIndex{9}\): We tin can use a bound-mass organisation to model a motorcycle suspension.

This organisation can be modeled using the aforementioned differential equation we used before:

\[mx″+bx′+kx=0. \nonumber\]

A motocross motorcycle weighs 204 lb, and we assume a rider weight of 180 lb. When the rider mounts the motorcycle, the pause compresses 4 in., so comes to residue at equilibrium. The break arrangement provides damping equal to 240 times the instantaneous vertical velocity of the motorcycle (and rider).

1. Prepare the differential equation that models the behavior of the motorcycle intermission system.
2. Nosotros are interested in what happens when the motorbike lands after taking a spring. Permit fourth dimension \[t=0\] denote the time when the motorbike beginning contacts the ground. If the motorbike hits the ground with a velocity of 10 ft/sec downward, find the equation of move of the motorcycle afterward the jump.
3. Graph the equation of move over the beginning second later on the motorcycle hits the ground.

Solution

1. We take defined equilibrium to be the point where \(mg=ks\), then we take

   \[\brainstorm{align*} mg &=ks \\ 384 &=k\left(\dfrac{one}{3}\correct)\\ one thousand &=1152. \stop{align*}\]

   We also have

   \[\begin{align*} West &=mg \\ 384 &=yard(32) \\ m &=12. \end{align*}\]

   Therefore, the differential equation that models the behavior of the motorcycle suspension is

   \[12x″+240x′+1152x=0. \nonumber\]

   Dividing through by 12, nosotros get

   \[x''+20x′+96x=0. \nonumber\]

2. The differential equation found in part a. has the full general solution

   \[x(t)=c_1e^{−8t}+c_2e^{−12t}. \nonumber \]

   Now, to make up one's mind our initial conditions, we consider the position and velocity of the motorcycle wheel when the wheel first contacts the ground. Since the motorcycle was in the air prior to contacting the ground, the wheel was hanging freely and the leap was uncompressed. Therefore the wheel is iv in. \(\left(\dfrac{1}{3}\text{ ft}\right)\) below the equilibrium position (with respect to the motorcycle frame), and nosotros have \(x(0)=\dfrac{1}{three}.\) According to the problem statement, the motorbike has a velocity of 10 ft/sec downward when the motorcycle contacts the ground, then \(x′(0)=10.\) Applying these initial weather condition, nosotros get \(c_1=\dfrac{seven}{2}\) and \(c_2=−\left(\dfrac{19}{six}\right)\),then the equation of motility is

   \[x(t)=\dfrac{7}{2}e^{−8t}−\dfrac{19}{six}eastward^{−12t}. \nonumber\]

3. The graph is shown in Figure \(\PageIndex{10}\).

Figure \(\PageIndex{10}\): Graph of the equation of motion over a time of one second.

LANDING VEHICLE

NASA is planning a mission to Mars. To save money, engineers have decided to adjust one of the moon landing vehicles for the new mission. Nonetheless, they are concerned nearly how the different gravitational forces will affect the suspension system that cushions the craft when information technology touches down. The acceleration resulting from gravity on the moon is one.6 m/sec², whereas on Mars it is three.7 chiliad/sec^two.

The suspension system on the craft can exist modeled as a damped spring-mass system. In this example, the leap is below the moon lander, and so the bound is slightly compressed at equilibrium, every bit shown in Figure \(\PageIndex{11}\).

Figure \(\PageIndex{11}\): The landing craft intermission tin can be represented equally a damped spring-mass arrangement. (credit "lander": NASA)

We retain the convention that downwardly is positive. Despite the new orientation, an exam of the forces affecting the lander shows that the same differential equation tin exist used to model the position of the landing craft relative to equilibrium:

\[mx''+bx′+kx=0, \nonumber\]

where \(g\) is the mass of the lander, \(b\) is the damping coefficient, and \(k\) is the spring constant.

1. The lander has a mass of 15,000 kg and the spring is 2 grand long when uncompressed. The lander is designed to shrink the spring 0.5 m to reach the equilibrium position under lunar gravity. The dashpot imparts a damping force equal to 48,000 times the instantaneous velocity of the lander. Set upward the differential equation that models the motion of the lander when the arts and crafts lands on the moon.
2. Allow fourth dimension \(t=0\) denote the instant the lander touches down. The rate of descent of the lander tin be controlled by the coiffure, so that it is descending at a rate of 2 m/sec when information technology touches downwardly. Find the equation of motion of the lander on the moon.
3. If the lander is traveling too fast when it touches down, it could fully shrink the spring and "bottom out." Bottoming out could damage the landing craft and must exist avoided at all costs. Graph the equation of movement found in part 2. If the spring is 0.five thou long when fully compressed, will the lander be in danger of bottoming out?
4. Assuming NASA engineers make no adjustments to the leap or the damper, how far does the lander compress the spring to reach the equilibrium position under Martian gravity?
5. If the lander crew uses the aforementioned procedures on Mars equally on the moon, and keeps the charge per unit of descent to two m/sec, will the lander lesser out when information technology lands on Mars?
6. What adjustments, if any, should the NASA engineers make to utilize the lander safely on Mars?

Forced Vibrations

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

\[mx″+bx′+kx=f(t), \nonumber\]

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

\[x(t)=c_1x_1(t)+c_2x_2(t)+x_p(t), \nonumber\]

where \(c_1x_1(t)+c_2x_2(t)\) is the general solution to the complementary equation and \(x_p(t)\) is a particular solution to the nonhomogeneous equation. If the arrangement is damped, \(\lim \limits_{t \to \infty} c_1x_1(t)+c_2x_2(t)=0.\) Since these terms do not impact the long-term behavior of the system, we call this function of the solution the transient solution . The long-term behavior of the system is determined by \(x_p(t)\), so we phone call this role of the solution the steady-country solution.

Case \(\PageIndex{7}\): Forced Vibrations

A mass of one slug stretches a spring 2 ft and comes to remainder at equilibrium. The system is attached to a dashpot that imparts a damping force equal to eight times the instantaneous velocity of the mass. Notice the equation of motion if an external forcefulness equal to \(f(t)=8 \sin (4t)\) 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

\[10″+8x′+16x=eight \sin (4t). \nonumber\]

The general solution to the complementary equation is

\[c_1e^{−4t}+c_2te^{−4t}. \nonumber\]

Assuming a item solution of the form \(x_p(t)=A \cos (4t)+ B \sin (4t)\) and using the method of undetermined coefficients, we detect \(x_p (t)=−\dfrac{one}{4} \cos (4t)\), so

\[x(t)=c_1e^{−4t}+c_2te^{−4t}−\dfrac{i}{iv} \cos (4t). \nonumber\]

At \(t=0,\) the mass is at rest in the equilibrium position, and then \(x(0)=ten′(0)=0.\) Applying these initial weather to solve for \(c_1\) and \(c_2,\) nosotros go

\[10(t)=\dfrac{one}{iv}e^{−4t}+te^{−4t}−\dfrac{1}{4} \cos (4t). \nonumber\]

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

Exercise \(\PageIndex{6}\)

A mass of ii kg is attached to a spring with constant 32 N/one thousand and comes to balance in the equilibrium position. Beginning at time\(t=0\), an external forcefulness equal to \(f(t)=68e^{−2}t \cos (4t) \) is applied to the organisation. Discover the equation of movement if at that place is no damping. What is the transient solution? What is the steady-state solution?

Hint

Observe the particular solution earlier applying the initial conditions.

Respond

\(x(t)=−\dfrac{i}{2} \cos (4t)+ \dfrac{9}{iv} \sin (4t)+ \dfrac{i}{2} e^{−2t} \cos (4t)−2e^{−2t} \sin (4t)\)

\(\text{Transient solution:} \dfrac{1}{ii}east^{−2t} \cos (4t)−2e^{−2t} \sin (4t)\)

\(\text{Steady-state solution:} −\dfrac{1}{2} \cos (4t)+ \dfrac{9}{4} \sin (4t) \)

RESONANCE

Consider an undamped organisation exhibiting simple harmonic movement. In the real earth, we never truly have an undamped system; –some damping always occurs. For theoretical purposes, however, we could imagine a spring-mass system independent in a vacuum chamber. With no air resistance, the mass would continue to movement upwardly and downwardly indefinitely.

The frequency of the resulting motion, given by \(f=\dfrac{1}{T}=\dfrac{ω}{2π}\), is called the natural frequency of the system. If an external force acting on the arrangement has a frequency close to the natural frequency of the system, a phenomenon called resonance results. The external force reinforces and amplifies the natural movement of the arrangement.

1. Consider the differential equation \(x″+x=0.\) Find the general solution. What is the natural frequency of the system?
2. Now suppose this system is subjected to an external strength given by \(f(t)=v \cos t.\) Solve the initial-value problem \(x″+x=v \cos t\), \(x(0)=0\), \(x′(0)=1\).
3. Graph the solution. What happens to the behavior of the organization over time?
4. In the existent globe, there is e'er some damping. Nevertheless, if the damping forcefulness is weak, and the external force is strong plenty, existent-world systems can yet showroom resonance. One of the most famous examples of resonance is the plummet of the Tacoma Narrows Bridge on November 7, 1940. The bridge had exhibited strange behavior ever since it was built. The roadway had a strange "bounce" to information technology. On the day it collapsed, a strong windstorm caused the roadway to twist and ripple violently. The bridge was unable to withstand these forces and it ultimately collapsed. Experts believe the windstorm exerted forces on the bridge that were very close to its natural frequency, and the resulting resonance ultimately shook the bridge apart.

   This website contains more information about the plummet of the Tacoma Narrows Span.

   During the short time the Tacoma Narrows Span stood, it became quite a tourist attraction. Several people were on site the solar day the bridge collapsed, and ane of them caught the plummet on film.

Watch the video to see the collapse of the Tacoma Narrows Bridge "Gallopin' Gertie". https://world wide web.youtube.com/lookout man?v=j-zczJXSxnw

1. Another existent-world example of resonance is a singer shattering a crystal wineglass when she sings simply the right note. When someone taps a crystal wineglass or wets a finger and runs it around the rim, a tone can be heard. That notation is created by the wineglass vibrating at its natural frequency. If a singer and then sings that same note at a loftier enough volume, the glass shatters as a result of resonance.

   The Tv show Mythbusters aired an episode on this phenomenon. Visit this website to learn more about information technology. Adam Savage likewise described the experience. Watch this video for his account.

The RLC Series Circuit

Consider an electric circuit containing a resistor, an inductor, and a capacitor, as shown in Figure \(\PageIndex{12}\). Such a circuit is called an RLC series circuit. RLC circuits are used in many electronic systems, most notably every bit tuners in AM/FM radios. The tuning knob varies the capacitance of the capacitor, which in plough tunes the radio. Such circuits can be modeled by second-club, constant-coefficient differential equations.

Let \(I(t)\) denote the current in the RLC circuit and \(q(t)\) denote the accuse on the capacitor. Furthermore, let \(L\) announce inductance in henrys (H), \(R\) denote resistance in ohms \((Ω)\), and \(C\) announce capacitance in farads (F). Last, let \(Due east(t)\) denote electrical potential in volts (V).

Kirchhoff's voltage dominion states that the sum of the voltage drops effectually any closed loop must be naught. 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 Figure \(\PageIndex{12}\) includes a voltage source, \(E(t)\), which adds voltage to the circuit, we take \(E_L+E_R+E_C=E(t)\).

Nosotros present the formulas beneath without further development and those of you interested in the derivation of these formulas can review the links. Using Faraday'south constabulary and Lenz's law, the voltage drop across an inductor tin exist shown to be proportional to the instantaneous charge per unit of change of current, with proportionality constant \(L.\) Thus,

\[E_L=L\dfrac{dI}{dt}. \nonumber\]

Adjacent, co-ordinate to Ohm's law, the voltage driblet across a resistor is proportional to the current passing through the resistor, with proportionality constant \(R.\) Therefore,

\[E_R=RI. \nonumber\]

Last, the voltage driblet across a capacitor is proportional to the charge, \(q,\) on the capacitor, with proportionality constant \(ane/C\). Thus,

\[E_C=\dfrac{one}{C}q. \nonumber\]

Adding these terms together, nosotros get

\[L\dfrac{dI}{dt}+RI+\dfrac{1}{C}q=E(t). \nonumber\]

Noting that \(I=(dq)/(dt)\), this becomes

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

Mathematically, this system is analogous to the spring-mass systems we have been examining in this section.

Figure \(\PageIndex{12}\): An RLC series circuit tin exist modeled by the same differential equation as a mass-leap system.

Series Circuit

Detect the accuse on the capacitor in an RLC series circuit where \(L=5/3\) H, \(R=10Ω\), \(C=i/xxx\) 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?

Solution

We accept

\[\begin{align*} L\dfrac{d^2q}{dt^2}+R\dfrac{dq}{dt}+\dfrac{ane}{C}q &=Eastward(t) \\[4pt] \dfrac{five}{3} \dfrac{d^2q}{dt^2}+10\dfrac{dq}{dt}+30q &=300 \\[4pt] \dfrac{d^2q}{dt^2}+half dozen\dfrac{dq}{dt}+18q &=180. \end{align*}\]

The full general solution to the complementary equation is

\[eastward^{−3t}(c_1 \cos (3t)+c_2 \sin (3t)). \nonumber\]

Assume a detail solution of the form \(q_p=A\), where \(A\) is a constant. Using the method of undetermined coefficients, we find \(A=ten\). And then,

\[q(t)=e^{−3t}(c_1 \cos (3t)+c_2 \sin (3t))+ten. \nonumber\]

Applying the initial weather \(q(0)=0\) and \(i(0)=((dq)/(dt))(0)=9,\) we find \(c_1=−10\) and \(c_2=−7.\) So the charge on the capacitor is

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

Looking closely at this function, we meet 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.

Exercise \(\PageIndex{vii}\)

Find the charge on the capacitor in an RLC series circuit where \(L=i/5\) H, \(R=2/5Ω,\) \(C=one/2\) F, and \(Eastward(t)=fifty\) 5. Assume the initial charge on the capacitor is 0 C and the initial electric current is 4 A.

Hint

Remember, \(E_L=L((dI)/(dt)).\)

Answer

\[q(t)=−25e^{−t} \cos (3t)−7e^{−t} \sin (3t)+25 \nonumber\]

Central Concepts

• 2nd-order constant-coefficient differential equations can be used to model bound-mass systems.
• An examination of the forces on a bound-mass system results in a differential equation of the class \[mx″+bx′+kx=f(t), \nonumber\] where mm represents the mass, bb is the coefficient of the damping force, \(k\) is the jump abiding, and \(f(t)\) represents whatever net external forces on the arrangement.
• If \(b=0\), at that place is no damping force acting on the organization, and simple harmonic motion results.
• If \(b≠0\),the behavior of the system depends on whether \(b^2−4mk>0, b^two−4mk=0,\) or \(b^2−4mk<0.\)
  • If \(b^2−4mk>0,\) the organisation 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 event in oscillatory beliefs.
  • If \(b^two−4mk<0\), the system is underdamped. It exhibits oscillatory beliefs, but the amplitude of the oscillations decreases over fourth dimension.
• If\(f(t)≠0\), the solution to the differential equation is the sum of a transient solution and a steady-country solution. The steady-land solution governs the long-term behavior of the system.
• The charge on the capacitor in an RLC serial circuit can also be modeled with a second-order constant-coefficient differential equation of the class \[Fifty\dfrac{d^2q}{dt^2}+R\dfrac{dq}{dt}+\dfrac{ane}{C}q=East(t), \nonumber\] where \(50\) is the inductance, \(R\) is the resistance, \(C\) is the capacitance, and \(Due east(t)\) is the voltage source.

Primal Equations

• Equation of simple harmonic move \[10″+ω^2x=0 \nonumber\]
• Solution for simple harmonic motion \[x(t)=c_1 \cos (ωt)+c_2 \sin (ωt) \nonumber\]
• Alternative form of solution for SHM \[x(t)=A \sin (ωt+ϕ) \nonumber\]
• Forced harmonic motion \[mx″+bx′+kx=f(t)\nonumber\]
• Charge in a RLC series circuit \[L\dfrac{d^2q}{dt^ii}+R\dfrac{dq}{dt}+\dfrac{i}{C}q=E(t),\nonumber\]

Glossary

RLC series circuit

a consummate electric path consisting of a resistor, an inductor, and a capacitor; a 2d-society, abiding-coefficient differential equation can be used to model the accuse on the capacitor in an RLC series circuit

elementary harmonic motion

motion described by the equation \(x(t)=c_1 \cos (ωt)+c_2 \sin (ωt)\), as exhibited by an undamped spring-mass arrangement in which the mass continues to oscillate indefinitely

steady-state solution

a solution to a nonhomogeneous differential equation related to the forcing role; in the long term, the solution approaches the steady-state solution

Contributors and Attributions

• Gilbert Strang (MIT) and Edwin "Jed" Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.

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Source: https://math.libretexts.org/Bookshelves/Calculus/Book%3A_Calculus_(OpenStax)/17%3A_Second-Order_Differential_Equations/17.3%3A_Applications_of_Second-Order_Differential_Equations

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