Download Second Order Equations: Understanding the Differences and Applications and more Summaries Differential Equations in PDF only on Docsity!
Chapter 2
Second Order Equations
2.1 Second Derivatives in Science and Engineering
Second order equations involve the second derivative d 2 y=dt^2. Often this is shortened to y^00 , and then the first derivative is y^0. In physical problems, y^0 can represent velocity v and the second derivative y^00 D a is acceleration : the rate dy 0 =dt that velocity is changing.
The most important equation in dynamics is Newton’s Second Law F D ma. Compare a second order equation to a first order equation, and allow them to be nonlinear :
First order y^0 D f .t; y/ Second order y 00 D F .t; y; y^0 / (1)
The second order equation needs two initial conditions , normally y.0/ and y^0 .0/— the initial velocity as well as the initial position. Then the equation tells us y 00 .0/ and the movement begins.
When you press the gas pedal, that produces acceleration. The brake pedal also brings acceleration but it is negative (the velocity decreases). The steering wheel produces acceleration too! Steering changes the direction of velocity, not the speed.
Right now we stay with straight line motion and one–dimensional problems :
d^2 y dt^2
0 (speeding up)
d^2 y dt^2
< 0 (slowing down):
The graph of y.t/ bends upwards for y^00 > 0 (the right word is convex ). Then the velocity y 0 (slope of the graph) is increasing. The graph bends downwards for y^00 < 0 ( concave ). Figure 2.1 shows the graph of y D sin t, when the acceleration is a D d 2 y=dt^2 D sin t. The important equation y^00 D y leads to sin t and cos t.
Notice how the velocity dy=dt (slope of the graph) changes sign in between zeros of y.
74 Chapter 2. Second Order Equations
y
v D (^0) 2 3 (^) t
y D sin t
y 00 D sin t
y^0 D cos t
a D y 00 > 0
v D y 0 < 0
y is going down and bending up
Figure 2.1: y^00 > 0 means that velocity y 0 (or slope) increases. The curve bends upward.
The best examples of F D ma come when the force F is ky, a constant k times the “position” or “displacement” y.t/. This produces the oscillation equation.
Fundamental equation of mechanics m d^2 y dt^2
C ky D 0 (2)
Think of a mass hanging at the bottom of a spring (Figure 2.2). The top of the spring is fixed, and the spring will stretch. Now stretch it a little more (move the mass downward by y.0/) and let go. The spring pulls back on the mass. Hooke’s Law says that the force is F D ky, proportional to the stretching distance y. Hooke’s constant is k. The mass will oscillate up and down. The oscillation goes on forever, because equation (2) does not include any friction (damping term b dy=dt). The oscillation is a perfect cosine, with y D cos !t and! D
p k=m, because the second derivative has to produce k=m to match y 00 D .k=m/y.
Oscillation at frequency! D
r k m
y D y.0/ cos
r k m
t
At time t D 0 , this shows the extra stretching y.0/. The derivative of cos !t has a factor ! D
p k=m. The second derivative y 00 has the required!^2 D k=m, so my 00 D ky. The movement of one spring and one mass is especially simple. There is only one fre- quency !. When we connect N masses by a line of springs there will be N frequencies— then Chapter 6 has to study the eigenvalues of N by N matrices.
m
d^2 y dt^2
D ky
.........................^ .... ................................................................ ................................................................. ................................................................ ............................................................................ ............. m y > 0 y^00 < 0 spring pulls up
y < 0 y^00 > 0 spring pushes down
y
........................................ ....................................................................................... ... ...................... ................... (^) ......................... .................... ............... m
Figure 2.2: Larger k D stiffer spring D faster !. Larger m D heavier mass D slower !.
76 Chapter 2. Second Order Equations
Frequency! or f
If the time t is measured in seconds , the frequency! is in radians per second. Then !t is in radians—it is an angle and cos !t is its cosine. But not everyone thinks naturally about radians. Complete cycles are easier to visualize. So frequency is also mea- sured in cycles per second. A typical frequency in your home is f D 60 cycles per second. One cycle per second is usually shortened to f D 1 Hertz. A complete cycle is 2 radians, so f D 60 Hertz is the same frequency as! D 120 radians per second. The period is the time T for one complete cycle. Thus T D 1=f. This is the only page where f is a frequency—on all other pages f .t/ is the driving function.
Frequency! D 2f Period T D
f
D
T D
f
D
y D A cos !t
D A cos
r k m
t f^ D^
! D
r k m
A A
t D 0 t D T time
Figure 2.3: Simple harmonic motion y D A cos !t : amplitude A and frequency !.
Harmonic Motion and Circular Motion
Harmonic motion is up and down (or side to side). When a point is in circular motion, its projections on the x and y axes are in harmonic motion. Those motions are closely related, which is why a piston going up and down can produce circular motion of a flywheel. The harmonic motion “speeds up in the middle and slows down at the ends” while the point moves with constant speed around the circle.
cos !t
sin !t^ ei!t
0 ! 2!
t
1 x D cos !t
0 ! 2!
t
y D sin !t
Figure 2.4: Steady motion around a circle produces cosine and sine motion along the axes.
2.1. Second Derivatives in Science and Engineering 77
Response Functions
I want to introduce some important words. The response is the output y.t/. Up to now the only inputs were the initial values y.0/ and y 0 .0/. In this case y.t/ would be the initial value response (but I have never seen those words). When we only see a few cycles of the motion, initial values make a big difference. In the long run, what counts is the response to a forcing function like f D cos !t. Now! is the driving frequency on the right hand side, where the natural frequency !n D
p k=m is decided by the left hand side :! comes from yp , !n comes from yn.
When the motion is driven by cos !t, a particular solution is yp D Y cos !t :
Forced motion yp.t/ at frequency! my 00 C ky D cos !t yp.t/ D
k m!^2
cos !t: (11)
To find yp .t/, I put Y cos !t into my 00 C ky and the result was .k m!^2 /Y cos !t. This matches the driving function cos !t when Y D 1=.k m!^2 /. The initial conditions are nowhere in equation (11). Those conditions contribute the null solution yn, which oscillates at the natural frequency !n D
p k=m. Then k D m!n^2. If I replace k by m!^2 n in the response yp .t/, I see!^2 n !^2 in the denominator :
Response to cos !t yp.t/ D
m
!^2 n !^2
cos !t. (12)
Our equation my 00 C ky D cos !t has no damping term. That will come in Section 2.3. It will produce a phase shift ˛. Damping will also reduce the amplitude jY.!/j. The amplitude is all we are seeing here in Y.!/ cos !t :
Frequency response Y.!/ D
k m!^2
D
m
!^2 n !^2
:^ (13)
The mass and spring, or the inductance and capacitance, decide the natural frequency !n. The response to a driving term cos !t (or ei !t^ ) is multiplication by the frequency response Y.!/. The formula changes when! D !n —we will study resonance!
With damping in Section 2.3, the frequency response Y.!/ will be a complex num- ber. We can’t escape complex arithmetic and we don’t want to. The magnitude jY.!/j will give the magnitude response (or amplitude response). The angle in the complex plane will decide the phase response (then ˛ D because we measure the phase lag).
The response is Y.!/ei !t^ to f .t/ D ei !t^ and the response is g.t/ to f .t/ D ı.t/. These show the frequency response Y from equation (13) and the impulse response g from equation (15). Yei !t^ and g.t/ are the two key solutions to my 00 C ky D f .t/.
2.1. Second Derivatives in Science and Engineering 79
REVIEW OF THE KEY IDEAS
1. my 00 Cky D 0 : A mass on a spring oscillates at the natural frequency !n D
p k=m.
2. my 00 C ky D cos !t : This driving force produces yp D .cos !t/=m
!n^2 !^2
3. There is resonance when !n D !. The solution yp D t sin !t includes a new factor t. 4. mg 00 Ckg D ı.t/ gives g.t/ D. sin !nt/=m!n D null solution with g 0 .0/ D 1=m. 5. Fundamental solution g : Every driving function f gives y.t/ D
R^ t 0
g.t s/f .s/ ds.
6. Frequency :! radians per second or f cycles per second (f Hertz). Period T D 1=f.
Problem Set 2.
1 Find a cosine and a sine that solve d 2 y=dt^2 D 9y. This is a second order equation so we expect two constants C and D (from integrating twice) :
Simple harmonic motion y.t/ D C cos !t C D sin !t: What is! ‹
If the system starts from rest (this means dy=dt D 0 at t D 0 ), which constant C or D will be zero?
2 In Problem 1, which C and D will give the starting values y.0/ D 0 and y^0 .0/ D 1?
3 Draw Figure 2.3 to show simple harmonic motion y D A cos .!t ˛/ with phases ˛ D =3 and ˛ D =2.
4 Suppose the circle in Figure 2.4 has radius 3 and circular frequency f D 60 Hertz. If the moving point starts at the angle 45 ı, find its x-coordinate A cos .!t ˛/. The phase lag is ˛ D 45 ı. When does the point first hit the x axis?
5 If you drive at 60 miles per hour on a circular track with radius R D 3 miles, what is the time T for one complete circuit? Your circular frequency is f D and your angular frequency is! D (with what units ?). The period is T.
6 The total energy E in the oscillating spring-mass system is
E D kinetic energy in mass C potential energy in spring D
m 2
dy dt
C
k 2
y^2 :
Compute E when y D C cos !t C D sin !t. The energy is constant!
7 Another way to show that the total energy E is constant :
Multiply my^00 C ky D 0 by y^0 : Then integrate my 0 y 00 and kyy 0 :
80 Chapter 2. Second Order Equations
8 A forced oscillation has another term in the equation and in the solution :
d 2 y dt^2
C 4y D F cos !t has y D C cos 2t C D sin 2t C A cos !t:
(a) Substitute y into the equation to see how C and D disappear (they give yn). Find the forced amplitude A in the particular solution yp D A cos !t. (b) In case! D 2 (forcing frequency D natural frequency), what answer does your formula give for A? The solution formula for y breaks down in this case.
9 Following Problem 8 , write down the complete solution yn C yp to the equation
m
d 2 y dt^2
C ky D F cos !t with! ¤ !n D
p k=m (no resonance):
The answer y has free constants C and D to match y.0/ and y^0 .0/ (A is fixed by F ).
10 Suppose Newton’s Law F D ma has the force F in the same direction as a :
my 00 D C ky including y 00 D 4y:
Find two possible choices of s in the exponential solutions y D est^. The solution is not sinusoidal and s is real and the oscillations are gone. Now y is unstable.
11 Here is a fourth order equation : d 4 y=dt^4 D 16y. Find four values of s that give exponential solutions y D est^. You could expect four initial conditions on y : y.0/ is given along with what three other conditions?
12 To find a particular solution to y 00 C 9y D ect^ , I would look for a multiple yp .t/ D Yect^ of the forcing function. What is that number Y? When does your formula give Y D 1? (Resonance needs a new formula for Y .)
13 In a particular solution y D Aei !t^ to y^00 C 9y D ei !t^ , what is the amplitude A? The formula blows up when the forcing frequency! D what natural frequency?
14 Equation (10) says that the tangent of the phase angle is tan ˛ D y^0 .0/=!y.0/. First, check that tan ˛ is dimensionless when y is in meters and time is in seconds. Next, if that ratio is tan ˛ D 1 , should you choose ˛ D =4 or ˛ D 5=4? Answer :
Separately you want R cos ˛ D y.0/ and R sin ˛ D y^0 .0/=!:
If those right hand sides are positive, choose the angle ˛ between 0 and =2. If those right hand sides are negative, add and choose ˛ D 5=4. Question : If y.0/ > 0 and y^0 .0/ < 0, does ˛ fall between =2 and or between 3=2 and 2? If you plot the vector from .0; 0/ to .y.0/; y 0 .0/=!/, its angle is ˛.
