Query 21

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course MTH 279

4/12 4:16 pm

Query 21 Differential Equations*********************************************

Question:  A 10 kg mass stretches a spring 9.8 cm beyond its original rest position. 

A driving force F(t) = 20 N * cos((8 s^-1) * t) begins at t = 0, where the downward direction is regarded as positive. 

Write down and solve the appropriate differential equation, obtaining the position function for the motion of the mass.

Plot your solution, and find the maximum distance of the mass from its equilibrium position.

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Your solution: 

 k = 10/0.098= 102 --> omega0 = sqrt (102/10) = 3.19

y_C= c1 sin 3.19t + c2 cos 3.19t

y_P= -F/(omega1^2-omega0^2) cos omega1*t

y_P = -2/(64-10.2) cos 8t= -0.037 cos 8t

y= c1 sin 3.19t + c2 cos 3.19t - 0.037 cos 8t

y(0)= 0 --> c2 = 0.037

y'(0) = 0 --> c1 =0

y= 0.037( cos 3.19t - cos 8t)

Max = 0.07

 

 

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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Self-critique rating:

@&

@& Good, but k = 1000 N / m, not 100 N/m.

Complete solution for comparison:

The net force on the spring is the sum of the spring tension and the driving force.

The spring tension is F_spring = - k y, where y is the displacement from equilibrium.

A 10 kg mass has weight 98 N, so the spring constant is k = 98 N / (9.8 cm) = 98 N / (.098 m) = 1000 N / m.

Thus

F_net = 20 cos( 8 t) - k y,

where we have suppressed the units but understand that all units are in Newtons and meters.

Our equation is therefore

m y '' = 20 cos(8 t) - k y,

which we rearrange to the form

y '' + k / m * y = 20 cos(8 t).

The homogeneous equation has fundamental set

{cos(omega t), sin(omega t)},

with omega = sqrt(k/m) = sqrt( 1000 / 10) = 10 (in units of rad / sec).

The complementary solution is therefore

y_C = A cos(10 t) + B sin(10 t).

We expect the particular solution to be a sum of multiples of cos(8 t) and sin(8 t):

y_P = A cos(8 t) + B sin( 8 t)

so that

y_P ' = -8 A sin(8 t) + 8 B cos(8 t)

and

y_P '' = -64 A cos(8 t) - 64 B sin(8 t).

Our equation

y '' + k / m * y = 20 cos(8 t)

becomes

-64 A cos (8 t) - 64 B sin (8 t) + 100 ( A cos(8 t) + B sin(8 t) ) = 20 cos(8 t).

Rearranging we get

36 A cos(8 t) + 36 B sin(8 t) = 20 cos(8 t)

so that

A = 0 and

B = 20 / 36 = 5/9.

Our particular solution is therefore

y = A cos(10 t) + B sin(10 t) + 5/9 cos(8 t).

Our initial conditions are that the mass is at equilibrium at t = 0, and stationary at that instant:

y(0) = 0

y ' (0) = 0.

y(0) = A + 5/9 = 0

so A = -5/9.

y ' (0) = 10 B + 40/9

so

B = -40 /9

and the equation of motion is

y = -5/9 cos(10 t) - 40/9 sin(10 t) + 5/9 cos(8 t).

The term -40/9 sin(10 t) is a sine function with period pi / 5 (approximately .63) and amplitude 40/9.

-5/9 cos(10 t) + 5/9 cos(8 t) will come in and out of phase with angular frequency 10 rad / sec - 8 rad / sec = 2 rad / sec, and will therefore do so with period pi.

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Question:  The motion of a mass is governed by the equation

m y '' + 2 delta y ' + omega_0^2 y = F(t),

with m = 2 kg, gamma = 8 kg / s and k = 80 N / m and F(t) = 20 N * e^(- t s^-1).

Solve the equation for the function y(t).

What is the long-term behavior of this system?

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Your solution: 

 r = -2 ± 6i

y_C= e^(-2t)(c1 sin 6t + c2 cos 6t)

y_P = Ae^-t

y'= -Ae^-t

y''= Ae^-t

Ae^-t + 4(-Ae^-t) + 40(Ae^-t) = 10e^-t

A - 4A + 40A = 10

A = 10/37 ≈ 0.27

y_P = 10/37*e^-t

y= e^(-2t)(c1 sin 6t + c2 cos 6t) + 0.27e^-t

y(0)=0 --> 0 = 1(0+c2) + 0.27*1 --> c2 = -0.27 = 10/37

y'= -2e^-2t(c1 sin 6t -0.27cos 6t)+ e^-2t((c1*6 cos 6t - 6*-.27 sin 6t) - 0.27e^-t

y'(0)= 0 --> -2(0-0.27) +1*(6*c1-0) - 0.27=0 --> c1= 0.045 = 5/111

y= e^(-2t)(-0.045sin 6t -0.27cos 6t) + 0.27e^-t

As t--> inf e^-2t and e^-t both --> 0 so y --> 0

It does this very quickly, within 5 seconds

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Given Solution: 

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Self-critique (if necessary):

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Self-critique rating:

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Question: 

Solve the equation

y '' + 2 delta y ' + omega_0^2 y = F cos( omega_1 * t), y(0) = 0, y ' (0) = 0.

Give an outline of your work.  A very similar problem was set up and partially solved in class on 110309, and your text gives the solution but not the steps.

Find the limiting function as omega_1 approaches omega_0, and discuss what this means in terms of a real system.

Find the limiting function as delta approaches 0, and discuss what this means in terms of a real system.

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Your solution:

For simplification: delta = d and omega = w so

y'' + 2d y'+ w0^2 = F cos w1*t 

r = -d ± sqrt w0^2-d^2)

y_C = e^(-delta*t)(c1sin( sqrt (w1^2-d^2)*t) + c2 cos( sqrt (w1^2-d^2)*t)

y_P= A sin w1*t + B cos w1*t

y'= A*w1*cos w1*t - B*w1*sin w1*t

y'' = -Aw1^2*sin w1*t - B*w1^2*cos w1*t

Sub into eq:

(-Aw1^2*sin w1*t - B*w1^2*cos w1*t) + 2d(A*w1*cos w1*t - B*w1*sin w1*t) +w0^2(A sin w1*t + B cos w1*t)

Simplify:

A(w0^2-w1^2) -B(2d*w1)= 0

B= A((w0^2-w1^2)/(2d*w1))

B(w0^2-w1^2) + A((2d*w1)) = F

Sub in for B:

A((w0^2-w1^2)/(2d*w1))*(w0^2-w1^2)+ A((2d*w1))= F

A = (F*(2d*w1))/((w0^2-w1^2)^2+(2d*w1)^2)

B= (F*(2d*w1))/((w0^2-w1^2)^2+(2d*w1)^2) * ((w0^2-w1^2)/(2d*w1))

B= (F(w0^2-w1^2))/((w0^2-w1^2)^2+(2d*w1)^2)

y_P= (F*(2d*w1))/((w0^2-w1^2)^2+(2d*w1)^2) * sin w1*t + (F(w0^2-w1^2))/((w0^2-w1^2)^2+(2d*w1)^2) * cos w1*t

y = e^(-delta*t)(c1 cos( sqrt (w1^2-d^2)*t) + c2 sin( sqrt (w1^2-d^2)*t) + (F*(2d*w1))/((w0^2-w1^2)^2+(2d*w1)^2) * sin w1*t + (F(w0^2-w1^2))/((w0^2-w1^2)^2+(2d*w1)^2) * cos w1*t

y(0)= 0 --> c2= -(F(w0^2-w1^2))/((w0^2-w1^2)^2+(2d*w1)^2)

y'(0)= 0 --> c1 = -(F*(2d*w1^2))/(((w0^2-w1^2)^2+(2d*w1)^2)* sqrt (w1^2-d^2))

y = (Fe^-dt)/((w0^2-w1^2)^2+(2d*w1)^2) * [-((2d*w1^2)/ sqrt (w1^2-d^2))* sin sqrt (w1^2-d^2)*t) - (w0^2-w1^2)*cos( sqrt (w1^2-d^2)*t)] + F/((w0^2-w1^2)^2+(2d*w1)^2) * [(2d*w1) sinw1*t + (w0^2-w1^2)cos w1*t]

This differs from the book by only the denominator of c1 where it has d(w0^2+w1^2) where I ended up with d*2w1^2.

w0= natural frequency of the system and w1 the frequency of the 'driver' and as the driver approaches w0 resonance occurs and A becomes indirectly proportional to delta. Also, the cosines drop out and that leaves only the sine for the undamped oscillation and then the e term is what diminishes that oscillation. As delta --> 0 the 'resistance' drops out and the system's frequency becomes the same as the driving frequency and it is the sines and e that 'drop out'.

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary): I don't know how I messed that c1 term up but as w1--> w0 the book's c1 denominator approaches mine.

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Self-critique rating:

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Question:  An LC circuit with L = 1 Henry and C = 4 microFarads is driven by voltage V_S(t) = 10 t e^(-t).

Write and solve the differential equation for the system.

Interpret your result.

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Your solution: 

 Q'' + 250,000Q = 10t*e^-t

Q_C = A sin 500t + B cos 500t

Q_P= Ate^-t + Be^-t

Q'= -Ate^-t + Ae^-t -Be^-t

Q''= Ate^-t -Ae^-t - A e^-t +Be^-t

(Ate^-t -2Ae^-t + Be^-t)+ 250,000(Ate^-t + Be^-t) = 10te^-t

A+250,000A = 10

 A = 10/250,001

-2A + B + 250,000 B = 0

B = 20/(250,001)^2

Q= A sin 500t + B cos 500t + 10/250,000t*e^-t + 20/(250,001)^2 e^-t

 There's no R here and Q doesn't fall off because of it since those 2 diminishing e terms are 'enhanced' by those very small coefficients to go to 0 very quickly and that's because the capacitance is very small compared to the inductance.

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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Self-critique rating:"

&#This looks good. See my notes. Let me know if you have any questions. &#