Sections 4.4 Written HW 3 due 10/24 or WebHW due 10/27 O ...gmarple/Oct20.pdf · MathLab o ce hour Sun 7-8 pm (MathLab) (Gary Marple) October 20th, 2017 Math 216: Introduction to

Agenda

Sections 4.4

Reminders

Written HW 3 due 10/24 or10/26

WebHW due 10/27

Office hours Tues, Thurs1-2 pm (5852 East Hall)

MathLab office hourSun 7-8 pm (MathLab)

(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations

§4.4 Mechanical and Electrical Vibrations

Objectives

Be able to write and solve the DE for an oscillating mass.

Know what it means if an oscillating mass isunderdamped, critically damped, or overdamped.

Be able to linearize the DE for a pendulum.


Example

A mass weighing 16 lb stretches a spring 3 in. The mass is setin motion from its equilibrium position with a downwardvelocity of 3 in/s. Assuming any damping force is negligible,find the position of the mass from its equilibrium position as afunction of time. Draw a phase portrait.


Newton’s second law.

Force = Mass · Acceleration

Let x represent the displacement of the mass from itsequilibrium point, k the spring constant, and m the mass ofthe object. The force due to the spring is given by Hooke’s lawwhich says that

Fs = −kx(t).

Summing up the forces gives

Fnet = mx ′′ = −kx ,

ormx ′′ + kx = 0.

Unforced, undamped oscillator





Fs = −kx(t).



ormx ′′ + kx = 0.






Fs = −kx(t).



ormx ′′ + kx = 0.






Fs = −kx(t).



ormx ′′ + kx = 0.



mx ′′ + kx = 0

Since the mass is initially at the equilibrium point withdownward velocity of 3 in/s, our initial conditions are

x(0) = 0, x ′(0) = 1/4.

At this point, we know all of the constants, except for thespring constant k . Recall that when the 16 lb weight wasadded to the spring, the spring stretched 3 inches. We can usethis to find the spring constant. After the mass was added tothe spring, all movement stopped meaning the system was inequilibrium. Therefore, the net force on the spring would havebeen zero. Using Hooke’s law, we have

0 = Fs + Fg ,

where Fg = mg is the force due to gravity.(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations

mx ′′ + kx = 0


x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg ,


mx ′′ + kx = 0


x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg ,


0 = Fs + Fg

In this case, we were given the weight of the mass, which is aforce. Therefore,

0 = −k/4 + 16, =⇒ k = 64 lb/ft.

Plugging in the constants gives the IVP

16

32x ′′ + 64x = 0, x(0) = 0, x ′(0) = 1/4

where we approximated the acceleration due to gravity asg ≈32 ft/s2. Simplifying gives

x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg


0 = −k/4 + 16, =⇒ k = 64 lb/ft.


16

32x ′′ + 64x = 0, x(0) = 0, x ′(0) = 1/4


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg


0 = −k/4 + 16, =⇒ k = 64 lb/ft.


16

32x ′′ + 64x = 0, x(0) = 0, x ′(0) = 1/4


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg


0 = −k/4 + 16, =⇒ k = 64 lb/ft.


16

32x ′′ + 64x = 0, x(0) = 0, x ′(0) = 1/4


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4

Recall that we’re looking for solutions of the form x = eλt .Plugging that in gives

λ2eλt + 128eλt = 0(λ2 + 128

)eλt = 0

Since eλt 6= 0 for all t, we must have

λ2 + 128 = 0 =⇒ λ = ±√−128 = ±i8

√2.

Therefore, the general solution is

x = c1 cos (8√

2t) + c2 sin (8√

2t).


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4


λ2eλt + 128eλt = 0(λ2 + 128

)eλt = 0


λ2 + 128 = 0 =⇒ λ = ±√−128 = ±i8

√2.


x = c1 cos (8√

2t) + c2 sin (8√

2t).


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4


λ2eλt + 128eλt = 0(λ2 + 128

)eλt = 0


λ2 + 128 = 0 =⇒ λ = ±√−128 = ±i8

√2.


x = c1 cos (8√

2t) + c2 sin (8√

2t).


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4


λ2eλt + 128eλt = 0(λ2 + 128

)eλt = 0


λ2 + 128 = 0 =⇒ λ = ±√−128 = ±i8

√2.


x = c1 cos (8√

2t) + c2 sin (8√

2t).


x ′′ + 128x = 0, x(0) = 0, x ′(0) = 1/4


λ2eλt + 128eλt = 0(λ2 + 128

)eλt = 0


λ2 + 128 = 0 =⇒ λ = ±√−128 = ±i8

√2.


x = c1 cos (8√

2t) + c2 sin (8√

2t).


x = c1 cos (8√

2t) + c2 sin (8√

2t)

We now need to enforce the initial conditions x(0) = 0 andx ′(0) = 1/4. Differentiating the general solution gives

x ′ = −8√

2c1 sin (8√

2t) + 8√

2c2 cos (8√

2).

We can now solve for c1 and c2.

0 = c1 cos 0 + c2 sin 0 =⇒ c1 = 0

1/4 = 8√

2c2 cos 0 =⇒ c2 = 1/(32√

2)

Therefore, the solution to the IVP is

x =sin (8

√2t)

32√

2.


x = c1 cos (8√

2t) + c2 sin (8√

2t)


x ′ = −8√

2c1 sin (8√

2t) + 8√

2c2 cos (8√

2).


0 = c1 cos 0 + c2 sin 0 =⇒ c1 = 0

1/4 = 8√

2c2 cos 0 =⇒ c2 = 1/(32√

2)


x =sin (8

√2t)

32√

2.


x = c1 cos (8√

2t) + c2 sin (8√

2t)


x ′ = −8√

2c1 sin (8√

2t) + 8√

2c2 cos (8√

2).


0 = c1 cos 0 + c2 sin 0 =⇒ c1 = 0

1/4 = 8√

2c2 cos 0 =⇒ c2 = 1/(32√

2)


x =sin (8

√2t)

32√

2.


x = c1 cos (8√

2t) + c2 sin (8√

2t)


x ′ = −8√

2c1 sin (8√

2t) + 8√

2c2 cos (8√

2).


0 = c1 cos 0 + c2 sin 0 =⇒ c1 = 0

1/4 = 8√

2c2 cos 0 =⇒ c2 = 1/(32√

2)


x =sin (8

√2t)

32√

2.


x = c1 cos (8√

2t) + c2 sin (8√

2t)


x ′ = −8√

2c1 sin (8√

2t) + 8√

2c2 cos (8√

2).


0 = c1 cos 0 + c2 sin 0 =⇒ c1 = 0

1/4 = 8√

2c2 cos 0 =⇒ c2 = 1/(32√

2)


x =sin (8

√2t)

32√

2.


x =sin (8

√2t)

32√

2


Let x1 = x and x2 = x ′. Then the phase portrait is given by

x1 =sin (8

√2t)

32√

2, x2 =

cos (8√

2t)

4.


Suppose we have an unforced, undamped oscillator. Recallthat the DE modeling the displacement of the mass from itsequilibrium position is given by

my ′′ + ky = 0

In the previous example, we found a solution that oscillates. Isthis always the case? What other types of behaviors might wesee? Let’s find out by solving the DE. We begin by finding theroots to the characteristic polynomial.

mλ2 + k = 0

λ1, λ2 = ±i√

k

m.

Since k and m are always positive, we only have one type ofmotion.



my ′′ + ky = 0


mλ2 + k = 0

λ1, λ2 = ±i√

k

m.




my ′′ + ky = 0


mλ2 + k = 0

λ1, λ2 = ±i√

k

m.




my ′′ + ky = 0


mλ2 + k = 0

λ1, λ2 = ±i√

k

m.



λ1, λ2 = ±i√

k

m

Simple Harmonic MotionAn unforced, undamped oscillator experiences simple harmonic motion.If we let ω2

0 = k/m, then the general solution to a DE of the form

y ′′ + ω20y = 0

isy = A cos (ω0t) + B sin (ω0t).

Sometimes it’s convenient to rewrite the solution in phase-amplitudeform

y = R cos (ω0t − δ),

where ω0 is called the natural frequency of vibration, R is theamplitude of motion, and δ is the phase. The period of the resultingmotion is

T =2π

ω0= 2π

√m

k.


λ1, λ2 = ±i√

k

m



y ′′ + ω20y = 0





T =2π

ω0= 2π

√m

k.


λ1, λ2 = ±i√

k

m



y ′′ + ω20y = 0





T =2π

ω0= 2π

√m

k.


λ1, λ2 = ±i√

k

m



y ′′ + ω20y = 0





T =2π

ω0= 2π

√m

k.


Example

Rewritey = 4 cos (3t)− 2 sin (3t) (?)

in phase-amplitude form

y = R cos (ω0t − δ).

Recall the trig identity

cos (a − b) = cos a cos b + sin a sin b.

If we multiply by R and let a = ω0t and b = δ, then

R cos (ω0t − δ) = R cos δ cosω0t + R sin δ sinω0t.

We can see from (?) that ω0 = 3, R cos δ = 4, andR sin δ = −2.


Example










Example










Example










R cos δ = 4 and R sin δ = −2

All that is left is to determine R and δ. Recall that the point (x , y)on a circle of radius r can be expressed in the form

x = r cos θ and y = r sin θ,

where θ is angle from the positive x-axis to the ray starting fromthe origin and passing through (x , y). If the point (4,−2) lies on acircle, the circle must have radius

R =√

42 + (−2)2 = 2√

5.

The point (4,−2) lies in the 4th quadrant, so the angle made withthe positive x-axis is

δ = 2π − arctan (2/4) ≈ 5.82 radians

Combining our results gives

y = 4 cos (3t)− 2 sin (3t) = 2√

5 cos (3t − (2π − arctan (1/2))) .






R =√

42 + (−2)2 = 2√

5.




y = 4 cos (3t)− 2 sin (3t) = 2√

5 cos (3t − (2π − arctan (1/2))) .






R =√

42 + (−2)2 = 2√

5.




y = 4 cos (3t)− 2 sin (3t) = 2√

5 cos (3t − (2π − arctan (1/2))) .






R =√

42 + (−2)2 = 2√

5.




y = 4 cos (3t)− 2 sin (3t) = 2√

5 cos (3t − (2π − arctan (1/2))) .






R =√

42 + (−2)2 = 2√

5.




y = 4 cos (3t)− 2 sin (3t) = 2√

5 cos (3t − (2π − arctan (1/2))) .


Example

A mass weighing 16 lb stretches a spring 3 in. The mass isattached to a viscous damper with a damping constant of 2lb·s/ft. The mass is set in motion from its equilibrium positionwith a downward velocity of 3 in/s. Find the position of themass from its equilibrium position as a function of time. Drawa phase portrait.


Recall Newton’s second law.


Let x represent the displacement of the mass from itsequilibrium point, γ the damping constant, k the springconstant, and m the mass of the object. The force due to thespring is given by Hooke’s law which says that

Fs = −kx(t).

The force due to damping is given by

Fd = −γx ′(t).


Fnet = mx ′′ = −kx − γx ′,or

mx ′′ + γx ′ + kx = 0.

Unforced, damped oscillator(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations




Fs = −kx(t).


Fd = −γx ′(t).



mx ′′ + γx ′ + kx = 0.





Fs = −kx(t).


Fd = −γx ′(t).



mx ′′ + γx ′ + kx = 0.





Fs = −kx(t).


Fd = −γx ′(t).



mx ′′ + γx ′ + kx = 0.





Fs = −kx(t).


Fd = −γx ′(t).



mx ′′ + γx ′ + kx = 0.





Fs = −kx(t).


Fd = −γx ′(t).



mx ′′ + γx ′ + kx = 0.


mx ′′ + γx ′ + kx = 0


x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg ,


mx ′′ + γx ′ + kx = 0


x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg ,


mx ′′ + γx ′ + kx = 0


x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg ,


0 = Fs + Fg


0 = −k/4 + 16, =⇒ k = 64 lb/ft.


16

32x ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg


0 = −k/4 + 16, =⇒ k = 64 lb/ft.


16

32x ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg


0 = −k/4 + 16, =⇒ k = 64 lb/ft.


16

32x ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


0 = Fs + Fg


0 = −k/4 + 16, =⇒ k = 64 lb/ft.


16

32x ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.

We know a second order linear DE with constant coefficientshas a solution of the form

x = eλt .

Plugging that into the DE gives

λ2eλt + 4λeλt + 128eλt = 0,(λ2 + 4λ + 128

)eλt = 0.

Since eλt 6= 0 for all t, λ must satisfy

λ2 + 4λ + 128 = 0 =⇒ λ = −2± i2√

31.


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t).


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


x = eλt .


λ2eλt + 4λeλt + 128eλt = 0,(λ2 + 4λ + 128

)eλt = 0.


λ2 + 4λ + 128 = 0 =⇒ λ = −2± i2√

31.


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t).


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


x = eλt .


λ2eλt + 4λeλt + 128eλt = 0,(λ2 + 4λ + 128

)eλt = 0.


λ2 + 4λ + 128 = 0 =⇒ λ = −2± i2√

31.


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t).


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


x = eλt .


λ2eλt + 4λeλt + 128eλt = 0,(λ2 + 4λ + 128

)eλt = 0.


λ2 + 4λ + 128 = 0 =⇒ λ = −2± i2√

31.


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t).


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


x = eλt .


λ2eλt + 4λeλt + 128eλt = 0,(λ2 + 4λ + 128

)eλt = 0.


λ2 + 4λ + 128 = 0 =⇒ λ = −2± i2√

31.


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t).


x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.


x = eλt .


λ2eλt + 4λeλt + 128eλt = 0,(λ2 + 4λ + 128

)eλt = 0.


λ2 + 4λ + 128 = 0 =⇒ λ = −2± i2√

31.


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t).


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t)

We now need to apply the initial conditions x(0) = 0 andx ′(0) = 1/4. Differentiating the general solution gives

x ′ =c1

[2e−2t cos (2

√31t)− 2

√31e−2t sin (2

√31t)

]+ c2

[2√

31e−2t cos (2√

31t)− 2e−2t sin (2√

31t)].

Plugging in the first initial condition into the general solutiongives

0 = c1e0 cos 0 + c2e

0 sin 0 =⇒ c1 = 0

Plugging c1 = 0 and the second initial condition into thederivative of the general solution gives

1/4 = c2

[2√

31e0 cos 0− 2e0 sin 0]

=⇒ c2 = 1/(8√

31)


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t)


x ′ =c1

[2e−2t cos (2

√31t)− 2

√31e−2t sin (2

√31t)

]+ c2

[2√

31e−2t cos (2√

31t)− 2e−2t sin (2√

31t)].


0 = c1e0 cos 0 + c2e

0 sin 0 =⇒ c1 = 0


1/4 = c2

[2√

31e0 cos 0− 2e0 sin 0]

=⇒ c2 = 1/(8√

31)


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t)


x ′ =c1

[2e−2t cos (2

√31t)− 2

√31e−2t sin (2

√31t)

]+ c2

[2√

31e−2t cos (2√

31t)− 2e−2t sin (2√

31t)].


0 = c1e0 cos 0 + c2e

0 sin 0 =⇒ c1 = 0


1/4 = c2

[2√

31e0 cos 0− 2e0 sin 0]

=⇒ c2 = 1/(8√

31)


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t)


x ′ =c1

[2e−2t cos (2

√31t)− 2

√31e−2t sin (2

√31t)

]+ c2

[2√

31e−2t cos (2√

31t)− 2e−2t sin (2√

31t)].


0 = c1e0 cos 0 + c2e

0 sin 0 =⇒ c1 = 0


1/4 = c2

[2√

31e0 cos 0− 2e0 sin 0]

=⇒ c2 = 1/(8√

31)


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t)


x ′ =c1

[2e−2t cos (2

√31t)− 2

√31e−2t sin (2

√31t)

]+ c2

[2√

31e−2t cos (2√

31t)− 2e−2t sin (2√

31t)].


0 = c1e0 cos 0 + c2e

0 sin 0 =⇒ c1 = 0


1/4 = c2

[2√

31e0 cos 0− 2e0 sin 0]

=⇒ c2 = 1/(8√

31)


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t)

Plugging c1 = 0 and c2 = 1/(8√

31) into the general solutiongives us the solution to the IVP.

x =1

8√

31e−2t sin (2

√31t)


x = c1e−2t cos (2

√31t) + c2e

−2t sin (2√

31t)

Plugging c1 = 0 and c2 = 1/(8√

31) into the general solutiongives us the solution to the IVP.

x =1

8√

31e−2t sin (2

√31t)


x =1

8√

31e−2t sin (2

√31t)


envelope =1

8√

31e−2t


Let x1 = x and x2 = x ′. Then the phase portrait is given by

x1 =1

8√

31e−2t sin (2

√31t), x2 = e−2t

(1

4cos (2

√31t)−

1

4√

31sin (2

√31t)

).


Suppose we have an unforced, damped oscillator. Recall thatthe DE modeling the displacement of mass from itsequilibrium position is given by

my ′′ + γy ′ + ky = 0


mλ2 + γλ + k = 0

λ1, λ2 =−γ ±

√γ2 − 4km

2m=

γ

2m

(−1±

√1− 4km

γ2

).

It turns out that there are three qualitatively differentsituations that can arise.



my ′′ + γy ′ + ky = 0


mλ2 + γλ + k = 0

λ1, λ2 =−γ ±

√γ2 − 4km

2m=

γ

2m

(−1±

√1− 4km

γ2

).




my ′′ + γy ′ + ky = 0


mλ2 + γλ + k = 0

λ1, λ2 =−γ ±

√γ2 − 4km

2m=

γ

2m

(−1±

√1− 4km

γ2

).




my ′′ + γy ′ + ky = 0


mλ2 + γλ + k = 0

λ1, λ2 =−γ ±

√γ2 − 4km

2m=

γ

2m

(−1±

√1− 4km

γ2

).



λ1, λ2 =γ

2m

(−1±

√1− 4km

γ2

).

1. Underdamped Harmonic Motion (γ2 − 4km < 0)In this case, the roots are complex numbers µ± iν withµ = −γ/(2m) < 0 and ν =

√4km − γ2/(2m) > 0. The general solution

has the formy = e−γt/2m(A cos νt + B sin νt).

2. Critically Damped Harmonic Motion (γ2 − 4km = 0)In this case, λ1 = −γ/(2m) < 0 is a repeated root. Therefore, thegeneral solution is

y = (A + Bt)e−γt/2m

3. Overdamped Harmonic Motion (γ2 − 4km > 0)Since m, γ, and k are positive, γ2 − 4km is always less than γ2.Therefore, the values of λ1 and λ2 are real, distinct, and negative. Thegeneral solution is

y = Aeλ1t + Beλ2t


λ1, λ2 =γ

2m

(−1±

√1− 4km

γ2

).







y = Aeλ1t + Beλ2t


λ1, λ2 =γ

2m

(−1±

√1− 4km

γ2

).







y = Aeλ1t + Beλ2t


λ1, λ2 =γ

2m

(−1±

√1− 4km

γ2

).







y = Aeλ1t + Beλ2t


Of the three cases, the most important is the first one,corresponding to underdamped harmonic motion. Recall thatthe solution looks like

y = e−γt/2m(A cos νt + B sin νt). (∗)

Sometimes it’s useful to rewrite the solution in a form that’seasy analyze. If we let A = R cos δ and B = R sin δ, then (∗)can be written in the form

y = Re−γt/2m cos (νt − δ).

where ν is called the quasi-frequency and Td = 2π/ν iscalled the quasi-period.


Of the three cases, the most important is the first one,corresponding to underdamped harmonic motion. Recall thatthe solution looks like

y = e−γt/2m(A cos νt + B sin νt). (∗)

Sometimes it’s useful to rewrite the solution in a form that’seasy analyze. If we let A = R cos δ and B = R sin δ, then (∗)can be written in the form

y = Re−γt/2m cos (νt − δ).

where ν is called the quasi-frequency and Td = 2π/ν iscalled the quasi-period.


So far, we’ve talked about two problem setups. The first is anunforced, undamped oscillator with DE

my ′′ + ky = 0.

The second is an unforced, damped oscillator with DE

my ′′ + γy ′ + ky = 0.

Now, suppose we have an external force F (t). This appears asan extra term on the right-hand side of the DE. So, for aforced, undamped oscillator, we would have

my ′′ + ky = F (t).

For a forced, damped oscillator, we would have

my ′′ + γy ′ + ky = F (t).

We will talk about these two cases in a later section.(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations


my ′′ + ky = 0.


my ′′ + γy ′ + ky = 0.


my ′′ + ky = F (t).


my ′′ + γy ′ + ky = F (t).



my ′′ + ky = 0.


my ′′ + γy ′ + ky = 0.


my ′′ + ky = F (t).


my ′′ + γy ′ + ky = F (t).



my ′′ + ky = 0.


my ′′ + γy ′ + ky = 0.


my ′′ + ky = F (t).


my ′′ + γy ′ + ky = F (t).


Example

(a) A pendulum of mass m swings along a circular path of radiusL. If the pendulum experiences a drag force that isproportional to its angular velocity, write a DE for the angle θbetween the rod and the downward vertical direction.

(b) Linearize the DE by assuming that sin θ ≈ θ when θ is small.


(a). We start by using Newton’s second law. In this case, wehave a force due to drag and a force due to gravity acting onthe pendulum. Therefore,

Fnet = mv ′ = −cθ′ −mg sin θ.

Since we need a DE in terms of θ, we need to express v , thetangential velocity, in terms of θ. Recall the identity

s = rθ

for the arc length s of a circle with radius r . Differentiatingboth sides with respect to t gives

v = rθ′.

In this case, the circular path has radius L. Therefore,

v = Lθ′.





s = rθ


v = rθ′.


v = Lθ′.





s = rθ


v = rθ′.


v = Lθ′.





s = rθ


v = rθ′.


v = Lθ′.


mv ′ = −cθ′ −mg sin θ.

Plugging v = Lθ into the DE gives

mLθ′′ = −cθ −mg sin θ

orθ′′ + γθ′ + ω2 sin θ = 0

where γ = c/mL and ω2 = g/L.





orθ′′ + γθ′ + ω2 sin θ = 0






orθ′′ + γθ′ + ω2 sin θ = 0



θ′′ + γθ′ + ω2 sin θ = 0

(b) To linearize the DE, we’re going to use the small angleapproximation which says that sin θ ≈ θ when θ is small. (Thisshould make sense because the first term of the Taylor seriesfor sin θ, centered at 0, is θ.) The linearized DE is

θ′′ + γθ′ + ω2θ = 0.


θ′′ + γθ′ + ω2 sin θ = 0

(b) To linearize the DE, we’re going to use the small angleapproximation which says that sin θ ≈ θ when θ is small. (Thisshould make sense because the first term of the Taylor seriesfor sin θ, centered at 0, is θ.) The linearized DE is

θ′′ + γθ′ + ω2θ = 0.


Example

The current i , measured in amperes, is a function of time t. Theresistance R (ohms), the capacitance C (farads), and the inductance L(henries) are all positive and are assumed to be known constants. Theimpressed voltage e (volts) is a given function of time. The total chargeq (coulombs) on the capacitor is related to the current by i = dq/dt.Kirchhoff’s voltage law states: in a closed circuit, the impressed voltageis equal to the algebraic sum of the voltages across the elements in therest of the circuit. If the voltage across the resistor is iR, the voltageacross the capacitor is q/C , and the voltage across the inductor isLdi/dt, write an IVP for the charge q on the capacitor.(Refer to the chalkboard.)


Sections 4.4 Written HW 3 due 10/24 or WebHW due 10/27 O ...gmarple/Oct20.pdf · MathLab o ce hour Sun 7-8 pm (MathLab) (Gary Marple) October 20th, 2017 Math 216: Introduction to

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