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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
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
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
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
x =sin (8
√2t)
32√
2
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
Let x1 = x and x2 = x ′. Then the phase portrait is given by
x1 =sin (8
√2t)
32√
2, x2 =
cos (8√
2t)
4.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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))) .
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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))) .
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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))) .
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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))) .
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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))) .
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
Recall Newton’s second law.
Force = Mass · Acceleration
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).
Summing up the forces gives
Fnet = mx ′′ = −kx − γx ′,or
mx ′′ + γx ′ + kx = 0.
Unforced, damped oscillator(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
Recall Newton’s second law.
Force = Mass · Acceleration
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).
Summing up the forces gives
Fnet = mx ′′ = −kx − γx ′,or
mx ′′ + γx ′ + kx = 0.
Unforced, damped oscillator(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
Recall Newton’s second law.
Force = Mass · Acceleration
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).
Summing up the forces gives
Fnet = mx ′′ = −kx − γx ′,or
mx ′′ + γx ′ + kx = 0.
Unforced, damped oscillator(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
Recall Newton’s second law.
Force = Mass · Acceleration
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).
Summing up the forces gives
Fnet = mx ′′ = −kx − γx ′,or
mx ′′ + γx ′ + kx = 0.
Unforced, damped oscillator(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
Recall Newton’s second law.
Force = Mass · Acceleration
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).
Summing up the forces gives
Fnet = mx ′′ = −kx − γx ′,or
mx ′′ + γx ′ + kx = 0.
Unforced, damped oscillator(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
Recall Newton’s second law.
Force = Mass · Acceleration
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).
Summing up the forces gives
Fnet = mx ′′ = −kx − γx ′,or
mx ′′ + γx ′ + kx = 0.
Unforced, damped oscillator(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
mx ′′ + γx ′ + 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 ′′ + γx ′ + 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 ′′ + γx ′ + 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
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 ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4
where we approximated the acceleration due to gravity asg ≈32 ft/s2. Simplifying gives
x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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 ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4
where we approximated the acceleration due to gravity asg ≈32 ft/s2. Simplifying gives
x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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 ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4
where we approximated the acceleration due to gravity asg ≈32 ft/s2. Simplifying gives
x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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 ′′ + 2x ′ + 64x = 0, x(0) = 0, x ′(0) = 1/4
where we approximated the acceleration due to gravity asg ≈32 ft/s2. Simplifying gives
x ′′ + 4x ′ + 128x = 0, x(0) = 0, x ′(0) = 1/4.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
Therefore, the general solution is
x = c1e−2t cos (2
√31t) + c2e
−2t sin (2√
31t).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
Therefore, the general solution is
x = c1e−2t cos (2
√31t) + c2e
−2t sin (2√
31t).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
Therefore, the general solution is
x = c1e−2t cos (2
√31t) + c2e
−2t sin (2√
31t).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
Therefore, the general solution is
x = c1e−2t cos (2
√31t) + c2e
−2t sin (2√
31t).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
Therefore, the general solution is
x = c1e−2t cos (2
√31t) + c2e
−2t sin (2√
31t).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
Therefore, the general solution is
x = c1e−2t cos (2
√31t) + c2e
−2t sin (2√
31t).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
x =1
8√
31e−2t sin (2
√31t)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
envelope =1
8√
31e−2t
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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)
).
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
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 + 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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
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 + 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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
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 + 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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
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 + 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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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
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
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
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
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
(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′.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
(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′.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
(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′.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
(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′.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations
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.)
(Gary Marple) October 20th, 2017 Math 216: Introduction to Differential Equations