Sections 3.7 and 3.8 Problems.ipynb

Kernel: SageMath 7.6

3.7 In each of Problems 1 and 2, determine $\omega$, $R$, and $\delta$ so as to write the given expression in the form $u=R\cos(\omega t-\delta)$.
#2. $u=-2\cos(\pi t)-3\sin(\pi t)$

ANS: $R=\sqrt{2^2+3^2}=\sqrt{13}, \omega=\pi$ and $\delta=\pi+\arctan(3/2)$ so that
$u=\sqrt{13}\cos(\pi t-(\pi+\arctan(3/2)))=\sqrt{13}(\cos(\pi t)\cos(\pi+\arctan(3/2))+\sin(\pi t)\sin(\pi+\arctan(3/2)))$
$=\sqrt{13}\left(\dfrac{-2}{\sqrt{13}}\cos(\pi t)-\dfrac{3}{\sqrt{13}}\sin(\pi t)\right)$
It's a help to draw the triangle,

from which it is evident that $\cos(\pi+\arctan(3/2))=\dfrac{-2}{\sqrt{13}}$ and $\sin(\pi+\arctan(3/2))=\dfrac{-3}{\sqrt{13}}$,
where the negatives come in because $\delta$ is in QIII.

Note that this solution would derive from an equation with roots $\pm i\pi$ and initial conditions $u(0)=-2$ and $u'(0)=-3$

http://www.bluffton.edu/homepages/facstaff/nesterd/java/slopefields.html

u,t=var('u t')
u = function('u')(t)
de = diff(u,t,2)+pi^2*u==0
desolve(de,u,[0,-2,-3])

-3*sin(pi*t)/pi - 2*cos(pi*t)

#6. A spring is stretched 10 cm by a force of 3 N. A mass of 2 kg
is hung from the spring and is also attached to a viscous damper that
exerts a force of 3 N when the velocity of the mass is 5 m/s. If the mass
is pulled down 5 cm below its equilibrium position and given an initial
downward velocity of 10 cm/s, determine its position $u$ at any time $t$.
Find the quasi-frequency $μ$ and the ratio of $μ$ to the natural frequency
of the corresponding undamped motion.

ANS: The spring constant satisfies $F=kx$, so $3=0.1k\Leftrightarrow k=30$.
$m=2$ and $\gamma 5=3\Leftrightarrow \gamma=0.6$, $u(0)=5, u'(0)=10$
Thus the equation from Newton's Second Law is
$$2u''+0.6u'+30u=0$$ The natural frequency is $\omega_0=\sqrt{\dfrac{k}{m}}=\sqrt{15}\approx 3.873$ while the quasi-frequency is
$\mu=\dfrac{\sqrt{4mk-\gamma^2}}{2m}=\dfrac{\sqrt{240-0.36}}{4}=\dfrac{\sqrt{239.64}}{4}=\dfrac{\sqrt{5991}}{20}\approx 3.870$

u,t=var('u t')
u = function('u')(t)
de = 2*diff(u,t,2)+0.6*diff(u,t)+30*u==0
desolve(de,u).show()

N(sqrt(5991)/20)

3.87007751860347

#8. A vibrating system satisfies the equation $u''+\gamma u'+u=0$. Find
the value of the damping coefficient $\gamma$ for which the quasi-period of
the damped motion is 50% greater than the period of the corresponding
undamped motion.

ANS: The natural (undamped) frequency is $\omega_0=\sqrt{1/1}=1$. The quasi-frequency is $\mu=\dfrac{\sqrt{4mk-\gamma^2}}{2m}=\dfrac{\sqrt{4-\gamma^2}}{2}=\sqrt{1-\left(\frac{\gamma}{2}\right)^2}$
We want the ratio of the quasi-period to the nature period to be $\dfrac{T_d}{T}=\dfrac{\omega_0}{\mu}=\dfrac{1}{\sqrt{1-\left(\frac{\gamma}{2}\right)^2}}=1.5$
Thus $1-\left(\frac{\gamma}{2}\right)^2=\dfrac{4}{9}\Rightarrow \gamma=\dfrac{2\sqrt{5}}{3}$

u,t=var('u t')
u = function('u')(t)
de1 = diff(u,t,2)+2*sqrt(5)*diff(u,t)/3+u==0
osc1=desolve(de1,u,[0,10,0])
osc1.show()
de2 = diff(u,t,2)+u==0
osc2=desolve(de2,u,[0,1,0])
osc2.show()
G=Graphics()
G+=plot(osc1,(t,0,4*pi))
G+=plot(osc2,(t,0,4*pi))
show(G)

It's a little hard to see, because the damping, but the damped oscillator seems to complete one "oscillation"
in the time it takes the undamped oscillator to complete one and a half.

#10. Show that the solution of the initial value problem
$$mu''+\gamma u'+ku=0, u(t_0)=u_0, u'(t_0)=u'_0$$ can be expressed as the sum $u=v+w$, where $v$ satisfies the initial
conditions $v(t_0)=u_0, v'(t_0)=0$, $w$ satisfies the initial conditions
$w(t_0)=0,w'(t_0)=u'_0$,and both $v$ and $w$ satisfy the same differential
equation as $u$. This is another instance of superposing solutions of
simpler problems to obtain the solution of a more general problem.

ANS: The general solution to the equation is $u(t)=v_1(t)+w_2(t)$ where $u(t_0)=v(t_0)+w(t_0)=u_0+0=u_0$
and $u'(t_0)=v'(t_0)+w'(t_0)=0+u'_0=u'_0$. Tada!

#12. If a series circuit has a capacitor of $C=0.8\times 10^{-6} F$ and an
inductor of $L=0.2 H$, find the resistance $R$ so that the circuit is
critically damped.

ANS: Recall the basic form for the ODE of an RCL circuit is $LQ''+RQ'+\dfrac{1}{C}Q=0$.
In this case, $0.2Q''+RQ'+\dfrac{Q}{0.8\times 10^{-6}}=0\Leftrightarrow Q''+5RQ'+2500^2Q=0$
The motion is critically damped if the discriminant of the characteristic polynomial is 0.
That is $R^2-4L/C=0\Leftarrow R=2\sqrt{L/C}$, in this case, $R=2\sqrt{(2500)^2}=5000$ or $5$k$\Omega$.

#16 To do this, we have to do problems 15 and 5. Here's problem 5:
#5. A mass of 20 g stretches a spring 5 cm. Suppose that the
mass is also attached to a viscous damper with a damping constant of
400 dyn·s/cm. If the mass is pulled down an additional 2 cm and then
released, find its position $u$ at any time $t$. Plot $u$ versus $t$. Determine the
quasi-frequency and the quasi-period. Determine the ratio of the quasi-
period to the period of the corresponding undamped motion. Also find
the time $\tau$ such that $|u(t)|<0.05$ cm for all $t>\tau$ .

ANS: The gravitational weight of 20 g = 0.02 kg is $0.02\cdot 9.81=0.1962$ N.
Hooke's law says there exists a constant $k$ such that $mg=kx\Rightarrow 0.1962=0.02k$ so $k=9.81$ kg/sec^2
The damping constant is given in terms of dyn·s/cm which in the mks system is N·s/m.
Because of the 100 to 1 ratio of cm to m, and the 1000 to 1 ratio of kg to g, the ratio of
Newtons to dynes is N/dyn = 10^5. So to convert from dyn·s/cm to N·s/m we multiply by $\dfrac{1 \text{N}}{10^5 \text{dyne}}\cdot\dfrac{10^2 \text{cm}}{\text{m}} = \dfrac{1}{10^3}$.
ParseError: KaTeX parse error: Undefined control sequence: \cdotp at position 1: \̲c̲d̲o̲t̲p̲ is then the damping coefficient in the mks system.

The equation of motion is then $0.02u''+0.4u'+9.81u=0$ and to form an Initial Value Problem we specify $u(0)=2$ and $u'(0)=0$

The characteristic polynomial is $0.02r^2+0.4r+9.81=0\Leftrightarrow r^2+20r+=(r+10)^2+490.5=0$

At this point I'm thinking to myself, self: Are you really committed to that 0.01 part of the acceleration due to gravity in the mks system?
And I hear the answer come wafting on the wind as if through a pine forest on the other side of the hill: "not so much..."
In that case, the equation is $(r+10)^2=-390\Leftrightarrow r=-10\pm \sqrt{390}i$,
meaning the solution is of the form $u(t)=e^{-10t}(c_1\cdot\cos(\sqrt{390}t)+c_2\cdot\sin(\sqrt{390}t))$
From the initial conditions, we find $u(0)=c_1=2$ and $u'(0)=-10(2)+\sqrt{390}c_2=0\Leftrightarrow c_2=\dfrac{2\sqrt{390}}{39}$

u,t=var('u t')
u = function('u')(t)
de = diff(u,t,2)+20*diff(u,t)+490*u==0
osc=desolve(de,u,[0,2,0])
osc.show()
G=Graphics()
G+=plot(osc,(t,0,1))
show(G)

The quasi-frequency is $\mu=\dfrac{\sqrt{4mk-\gamma^2}}{2m}=\dfrac{\sqrt{4(0.02)(9.8)-0.4^2}}{2(0.02)}\approx 19.748$
and the quasi-period is $T_d=\dfrac{2\pi}{\mu}\approx \dfrac{2\pi}{19.748}\approx 0.318$
The ratio of the quasi-period to the period of the corresponding undamped motion is
$\approx{19.75}\sqrt{\dfrac{0.02}{9.8}}\approx 0.892$, or, using the approximation,
$\dfrac{\mu}{\omega_0}\approx 1-\dfrac{\gamma^2}{8km}\approx 1-\dfrac{0.4^2}{8(9.8)(0.02)}\approx 0.898$

The time $\tau$ such that $|u(t)|<0.05$ cm for all $t>\tau$ is the solution to $\frac{2}{39} \, {\left(\sqrt{390} \sin\left(\sqrt{390} t\right) + 39 \, \cos\left(\sqrt{390} t\right)\right)} e^{\left(-10 \, t\right)}=0.05$
in the interval $(0.3,0.4)$. I'll use Sage's find_root command:fo

t = var('t')
find_root((2*(sqrt(390)*sin(sqrt(390)*t))/39+2*cos(sqrt(390)*t))*exp(-10*t)==0.05,0.3,0.4)

0.3672282167109013

That looks about right?

#15 Logarithmic Decrement.
a. For the damped oscillation described by $Re^{-\gamma t/(2m)}\cos(\mu t+\delta)$,
show that the time between successive maxima is $T_d=2\pi/\mu$.

ANS: The derivative of this function is $-Re^{-\gamma t/(2m)}\left(\dfrac{\gamma}{2m}\cos(\mu t+\delta)+\mu\sin(\mu t+\delta)\right)$
is also a damped oscillator with period $\dfrac{2\pi}{\mu}$

b. Show that the ratio of the displacements at two successive
maxima is given by $\exp\left(\dfrac{\gamma T_d}{2m}\right)=\exp\left(\dfrac{\gamma\pi}{m\mu}\right)$.
Observe that this ratio does not depend on which pair of maxima is chosen.
The natural logarithm of this ratio is called the logarithmic decrement and
is denoted by $\Delta$.

ANS: Note that $exp(\gamma T_d/(2m))=\exp(\gamma\pi/(m\mu))$
$u'(t)=-Re^{-\gamma t/(2m)}\left(\dfrac{\gamma}{2m}\cos(\mu t+\delta)+\mu\sin(\mu t+\delta)\right)=$ $-Re^{-\gamma t/(2m)}$ $\sqrt{\dfrac{\gamma^2}{4m^2}+\mu^2}\cos(\mu t+\psi)=0$,
where $\tan\psi=\dfrac{2\mu m}{\gamma}$, has zeros where $t=\dfrac{(2k+1)\pi}{2\mu}-\dfrac{2m}{\gamma}$
The ratio in question is $\dfrac{u(t_k)}{u(t_{k+1})}=\exp\left(\dfrac{-\gamma t_k}{2m}\right)/\exp\left(\dfrac{−\gamma t_{k+1}}{2m}\right)=\exp\left(\dfrac{\gamma(t_{k+1}−t_k)}{2m}\right)$.
The difference in times between peaks is the quasi-period: $t_{k+1}-t_k=\dfrac{2\pi}{\mu}$, so $\dfrac{u(t_k)}{u(t_{k+1})}=\exp\left(\dfrac{\gamma(2\pi /\mu)}{2m}\right)=\exp\left(\dfrac{\gamma T_d}{2m}\right)$

R,gamma,m,delta,mu,t=var('R gamma m delta mu t')
u=function('u')(t)
u=R*exp(-gamma*t/(2*m))*cos(mu*t+delta)
uprime=diff(u,t)
uprime
solve(uprime==0,t)

[sin(mu*t + delta) == -1/2*gamma*cos(mu*t + delta)/(m*mu)]

c. Show that $\Delta=\dfrac{\pi\gamma}{m\mu}$.
Since $m,\mu$, and $\Delta$ are quantities that can be measured easily
for a mechanical system, this result provides a convenient and practical
method for determining the damping constant of the system, which is more
difficult to measure directly. In particular, for the motion of a vibrating
mass in a viscous fluid, the damping constant depends on the viscosity of the
fluid; for simple geometric shapes the form of this dependence is known, and
the preceding relation allows the experimental determination of the viscosity.
This is one of the most accurate ways of determining the viscosity of a gas at high
pressure.

ANS: This part follows directly from the definition of $\ln$

#16. Referring to Problem 15, find the logarithmic decrement of the system in Problem 5.

$\Delta=\dfrac{\pi\gamma}{m\mu}\approx\dfrac{0.4\pi}{0.02(19.748)}\approx 3.226$
Note that $\exp(3.226)\approx 25.17$ is the ratio between peaks.

18. Consider the initial value problem $$mu''+\gamma u'+ku=0; u(0)=u_0, u'(0)=v_0.$$ Assume that $\gamma^2<4km$.
    a. Solve the initial value problem.
    ANS: $u(t)=e^{-\gamma t/(2m)}\left(c_1\cos\left(\dfrac{\sqrt{4mk-\gamma^2}}{2m}t\right)+c_2\sin\left(\dfrac{\sqrt{4mk-\gamma^2}}{2m}t\right)\right)$
    $u(0)=c_1=u_0$ and $u'(0)=-\dfrac{\gamma u_0}{2m}+\dfrac{\sqrt{4mk-\gamma^2}}{2m}c_2=v_0\Leftrightarrow c_2=\dfrac{2mv_0+\gamma u_0}{\sqrt{4m-\gamma^2}}$
    So $u(t)=e^{-\gamma t/(2m)}\left(u_0\cos\left(\dfrac{\sqrt{4mk-\gamma^2}}{2m}t\right)+\left(\dfrac{2mv_0+\gamma u_0}{\sqrt{4m-\gamma^2}}\right)\sin\left(\dfrac{\sqrt{4mk-\gamma^2}}{2m}t\right)\right)$
    
    

b. Write the solution in the form $u(t)=Re^{-\gamma t/2m}\cos(\mu t−\delta)$.
Determine $R$ in terms of $m,\gamma,k,u_0,$ and $v_0$.

ANS: $R=\sqrt{u_0^2+\dfrac{(2mv_0-\gamma u_0)^2}{4mk-\gamma^2}}$ and $\delta=\arctan\left(\dfrac{2mv_0-\gamma u_0}{u_0\sqrt{4mk-\gamma^2}}\right)$

c. Investigate the dependence of $R$ on the damping coefficient $\gamma$ for fixed values of the other parameters.

ANS: $\dfrac{d}{d\gamma}R=\left(u_0^2+\dfrac{(2mv_0-\gamma u_0)^2}{4mk-\gamma^2}\right)^{-1/2}\left(\dfrac{1}{\sqrt{4km-\gamma^2}}+\dfrac{\left(\gamma u_0-2\,mv_0\right)\gamma}{(4\,km-\gamma^2)^{3/2}u_0}\right)=0\Leftrightarrow$
$\dfrac{1}{\sqrt{4km-\gamma^2}}+\dfrac{\left(\gamma u_0-2\,mv_0\right)\gamma}{(4\,km-\gamma^2)^{3/2}u_0}=0$
After some arduous algebra, $\gamma = \dfrac{2ku_0}{v_0}$

R,gamma,m,k,delta,mu,t,u0,v0=var('R gamma m k delta mu t u0 v0')
R=function('R')(gamma)
R=sqrt(u0^2+(2*m*v0-gamma*u0)/(u0*sqrt(4*m*k-gamma^2)))
Rprime = diff(R,gamma)
Rprime.show()
solve(Rprime==0,gamma)

[gamma == 2*k*u0/v0]

#20. The position of a certain undamped spring-mass system satisfies
the initial value problem $$u''+2u=0; u(0)=0, u'(0)=2.$$ a. Find the solution of this initial value problem.

$u(t)=\sqrt{2}\sin{\sqrt{2}\,t}$

b. Plot $u$ versus $t$ and $u'$ versus $t$ on the same axes.

t,u=var('t u')
u=function('u')(t)
u=desolve(diff(u,t,2)+2*u==0,u,[0,0,2])
uprime(t)=diff(u,t)
G=Graphics()
G+=plot(u,(t,0,7))
G+=plot(diff(u,t),(t,0,7),linestyle="dotted",color=(0.3,0.18,0.2018))
G.show()

c. Plot $u'$ versus $u$; that is, plot $u(t)$ and $u'(t)$ parametrically
with $t$ as the parameter. This plot is known as a phase plot, and
the $u$-$u'$-plane is called the phase plane. Observe that a closed
curve in the phase plane corresponds to a periodic solution $u(t)$.
What is the direction of motion on the phase plot as $t$ increases?

ANS: Fortunately, sagemath/ipythonb has just the tool for this.
Note that, since $u$ is increasing initially, the phaseplot will
start from $(0,2)$ and go around in the clockwise direction.

parametric_plot( (u, uprime), (t, 0, 2*pi))

#22. In the absence of damping, the motion of a spring-mass system
satisfies the initial value problem $$mu''+ku=0; u(0)=a, u'(0)=b$$ a. Show that the kinetic energy initially imparted to the mass
is $mb^2/2$ and that the potential energy initially stored in the
spring is $ka^2/2$, so initially the total energy in the system is
$(ka^2 + mb^2)/2$.

ANS: This follows immediately from the definition of kinetic energy as $mv^2/2$ where $v_0=b$
and the potential energy while the potential energy comes entirely from the spring's perspective:
"How far am I from equilibrium?" If the spring's displacement from equlibrium is $u$, then, by Hooke's law
the force the spring exerts is $F=ku$ and the total potential energy is the work it takes to move a mass from equilibrium to a displacement $a$, found by integrating $\int\limits_0^a\,ks\,ds=\dfrac{k\,a^2}{2}$

b. Solve the given initial value problem.

ANS: c. Using the solution in part b, determine the total energy in the system at any time $t$. Your result should confirm the principle of conservation of energy for this system.

3.8#4. A mass of 5 kg stretches a spring 0.1 m. The mass
is acted on by an external force of $10\sin(t/2)$ N (newtons)
and moves in a medium that imparts a viscous force of 2 N
when the speed of the mass is 0.04 m/s. If the mass is set in motion from
its equilibrium position with an initial velocity of 0.03 m/s, formulate
the initial value problem describing the motion of the mass.

ANS: We can determine the spring constant from $kx=mg\Leftrightarrow 0.1k=5(9.8)\Leftrightarrow k=490$
And the damping constant is satisfies $0.04\gamma=2\Leftrightarrow \gamma=50$ so the equation of motion is
$5u''+50u'+490u=0\Leftrightarrow u''+10u'+98=0$ and the initial conditions are $u(0)=0$ and $u'(0)=0.03$

5. a. Find the solution of the initial value problem in Problem 4.
   

The characteristic equation $r^2+10r+98=0\Leftrightarrow (r+5)^2=-98+25\Leftrightarrow r=-5\pm\sqrt{73}i$
$u=e^{-5t}(c_1\cos\sqrt{73}t+c_2\sin\sqrt{73}t)$

A,B,t,u,c1,c2=var('A B t u c1 c2')
solve([391*A/4+5*B==0,-5*A+391*B/4==2],A,B)

[[A == (-160/153281), B == (3128/153281)]]

u=function('u')(t)
u=exp(-5*t)*(160*cos(sqrt(73)*t)/153281-764*sqrt(73)*sin(sqrt(73)*t))/(153281*73)+-160/153281*cos(t/2)+3128/153281*sin(t/2)
#solve([u(t).subs(t=0)==0,diff(u,t).subs(t=0)==0.03],c1,c2)
#u(t).subs(A=-160/153281, B=3128/153281, t=0)
u

-4/1715139742153*(29276671*sqrt(73)*sin(sqrt(73)*t) - 40*cos(sqrt(73)*t))*e^(-5*t) - 160/153281*cos(1/2*t) + 3128/153281*sin(1/2*t)

de=diff(u,t,2)+10*diff(u,t)+98*u==2*sin(t/2)
desolve(diff(u,t,2)+10*diff(u,t)+98*u==2*sin(t/2),u,[0,0,0.03],t)

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-22-e52f25ec1cc5> in <module>()
      1 de=diff(u,t,Integer(2))+Integer(10)*diff(u,t)+Integer(98)*u==Integer(2)*sin(t/Integer(2))
----> 2 desolve(diff(u,t,Integer(2))+Integer(10)*diff(u,t)+Integer(98)*u==Integer(2)*sin(t/Integer(2)),u,[Integer(0),Integer(0),RealNumber('0.03')],t)

/ext/sage/sage-7.6/local/lib/python2.7/site-packages/sage/calculus/desolvers.pyc in desolve(de, dvar, ics, ivar, show_method, contrib_ode)
    445     de00 = de._maxima_()
    446     P = de00.parent()
--> 447     dvar_str=P(dvar.operator()).str()
    448     ivar_str=P(ivar).str()
    449     de00 = de00.str()
/ext/sage/sage-7.6/local/lib/python2.7/site-packages/sage/interfaces/interface.pyc in __call__(self, x, name)
    256                 return cls(self, str(x), name=name)
    257             except TypeError:
--> 258                 raise TypeError(msg)
    259 
    260     def _coerce_from_special_method(self, x):
TypeError: unable to coerce element into maxima_lib

desolve?

b. Identify the transient and steady-state parts of the solution. G c. Plot the graph of the steady-state solution. N d. If the given external force is replaced by a force of 2 cos(ωt) of frequency ω, find the value of ω for which the amplitude of the forced response is maximum.

t, u = var('t u')
u=function('u')(t)
de = diff(u,t,2)+64*u==32*sin(8*t)
u = desolve(de,u,[0,1/4,0])
u.show()

plot(diff(u,t),(t,0,pi))

uprime = diff(u,t)
uprime

16*t*sin(8*t) - 2*sin(8*t)

solve(uprime==0,t)

[t == 0, t == (1/8)]

Using Variation of Parameters (no credit for other methods!), first find a particular solution of
$y''(t)−5y'(t)+4y(t)=3e^{4t}$
then use it to find the general solution, and finally, use the latter to solve the initial value problem $y''(t)−5y'(t)+4y(t)=3e^{4t}; y(0)=0,y_0(0)=1.$