Query 18

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course MTH 279

Query 18 Differential Equations*********************************************

Question: A 10 kg mass stretches a spring 30 mm beyond its unloaded position. The spring is pulled down to a position 70 mm below its unloaded position and released.

Write and solve the differential equation for its motion.

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Your solution:

w = mg = 98.1 kgm/s^2 = 98.1 N, F = -ky = -130.8 N

w+F = mg-kY = 0; Y = mg/k

0.03 = (98.1/k)

k = 3270 N/m

m*d^2/dt^2(Y+y) = w+F = mg-k(Y-y)

Simplifying:

my” + ky = 0

y” + ky/m = 0

ω^2 = k/m = 327 s^-2

y” + ω^2y = 0

Our characteristic polynomial is:

λ^2 + ω^2 = 0.

Using the quadratic formula:

λ = (0±sqrt(0-4*ω^2))/2 = ±i*sqrt(ω^2)

y_1 = e^(-i*sqrt(ω^2)*t)

y_2 = e^(i*sqrt(ω^2)*t)

Using Euler’s formula for complex exponentials:

y_1 = cos(sqrt(ω^2)t)-i*sin(sqrt(ω^2)t)

y_2 = cos(sqrt(ω^2)t) + i*sin(sqrt(ω^2)t)

We check if these are a fundamental set:

W(t) = |[y_1,y_2;y_1’,y_2’]| = sqrt(ω^2)*i - -sqrt(ω^2)*i = 2sqrt(ω^2)*i

This solution is non-zero for all values of t, so the solutions form a fundamental set.

y_g = y_1 + y_2= 2Ccos(sqrt(ω^2)t)

Imposing our initial condition:

y(0) = 0.04 = 2Ccos(0)

C = 0.02

y(t) = 0.04cos(sqrt(ω^2)t)

Plugging in values: ω^2 = k/m = 327s^(-2)

sqrt(ω^2) = 18.08 s^(-1)

y(t) = 0.04cos(18.08t)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary): OK

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Self-critique rating: OK

Question: The graph below represents position vs. clock time for a mass m oscillating on a spring with force constant k. The position function is y = R cos(omega t - delta).

Find delta, omega and R.

Give the initial conditions on the y and y '.

Determine the mass and the force constant.

Describe how you would achieve these initial conditions, given the appropriate mass and spring in front of you.

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Your solution:

Using basic knowledge of sinusoidal functions:

We know that R = peak amplitude, so R = 3.

ω=angular frequency= 2πf = 2π/T. Our period is 2, so our ω = pi.

The function is shifted to the right ¼, which means that δ = π/4.

y(t) = 3cos(πt-π/4) = 3cos(π(t-1/4))

ω^2 = k/m,

so k = m*ω^2 = π^2*m N/m.

There is not enough information to determine both the mass and spring coefficient.

Initial conditions: y(0) = 3/sqrt(2), y’(0) = 3π/sqrt(2) ≈ 6.66

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

I’m not sure if this is the correct way to do this problem, or if it’s the way you’d like it done, but I’m pretty sure it’s correct, or very nearly. The graph comes out nearly the same as the one given when plotted.

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Self-critique rating:3

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Question: This problem isn't in your text, but is related to the first problem in this set, and to one of the problems that does appear in your text. Repeating the first problem:

'A 10 kg mass stretches a spring 30 mm beyond its unloaded position. The spring is pulled down to a position 70 mm below its unloaded position and released.

Write and solve the differential equation for its motion.'

Suppose that instead of simply releasing it from its 70 mm position, you deliver a sharp blow to get it moving at downward at 40 mm / second.

What initial conditions apply to this situation?

Apply the initial conditions to the general solution of the differential equation, and give the resulting function.

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Your solution:

y(0) = 0 m, y’(0) = 0.04 m/s

y_g = Acos(sqrt(k/m)t)+Bsin(sqrt(k/m)t)

y_g’ = -sqrt(k/m)Asin(sqrt(k/m)t)+sqrt(k/m)Bcos(sqrt(k/m)t)

y(0) = 0 = A

A = 0

y’(0) = sqrt(k/m)*B = 0.04

B = 0.04/sqrt(k/m)

y(t) = (0.04/sqrt(k/m))*sin(sqrt(k/m)t)

Inserting values gives:

y(t) = (0.04/18.083)*sin(18.083t) = 0.002212sin(18.083t)

@&

Recall that the equilibrium position for the oscillator is 40 mm below its unloaded position.

The 70 mm position is 30 mm below equilibrium, so that initially y = -30 mm, giving us the initial condition

y(0) = -30 mm.

The initial velocity is 40 mm / sec downward, so that

y ' (0) = -40 mm/sec.

As before the equation of motion is

y '' = -k/m * y.

and its general solution is

y(t) = A cos(omega t) + B sin(omega t)

with omega = sqrt(k / m).

y(0) = -30 mm yields A = -30 mm.

y ' (t) = -omega A sin(omega t) + omega B cos(omega t)

so that

y ' (0) = omega * B.

Recall that omega = 18 rad / sec, approx., so

y ' (0) = 18 rad / sec * B = -40 mm / sec

so that

B = -40 mm/s / (18 rad /s) = -2.3 mm.

The equation of motion for the mass is therefore

y(t) = -30 cos( 18 t) - 2.3 sin(18 t).

*@

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating: OK

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Question: The 32-pound weight at its equilibrium position in a critically-damped spring-and-dashpot system is given a 4 ft / sec downward initial velocity, and attains a maximum displacement of 6 inches from equilibrium. What are the values of the drag force constant gamma and the spring constant k?

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Your solution:

F = ma

32lbf = 32m

m = 1 slug

γ^2 = 4km

λ = -γ/(2m)

y_1 = e^(λt)

y_2 = te^(λt)

y_g = Ae^(λt) + Bte^(λt)

y_g’ = Aλe^(λt) + Be^(λt) + Btλe^(λt)

y’(0) = 4 = Aλ + B

A = (4-B)/λ

y(0) = 0 = (4-B)/γ

B = 4

A = 0

y(t) = 4te^(λt)

4te^(λt) = 6

4e^(λt) + 4tλe^(λt) = 0

λ = -2/3*e^(-1), t = 3e/2

y(t) = 4te^(-2/3e^(-1)t)

γ^2 = 4km

λ = -γ/(2) = -2/3e^(-1)

γ = 4e^(-1)/3 ≈ 0.4905

k = γ^2/4 = 4e^(-2)/9 ≈ 0.0601 lbf/ft

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: The motion of a system is governed by the equation m y '' + gamma y ' + k y = 0, with y(0) = 0 and y ' (0) = v_0.

Give the solutions which correspond to the critically damped, overdamped and underdamped cases.

Show that as gamma approaches 2 sqrt(k * m) from above, the solution to the underdamped case approaches the solution to the critically damped case.

Show that as gamma approaches 2 sqrt(k * m) from below, the solution to the overdamped case approaches the solution to the critically damped case.

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Your solution:

Critically Damped:

λ = -γ/2m

y_g = Ae^(λt) + Bte^(λt)

y_g’ = Aλe^(λt) + Be^(λt) + Btλe^(λt)

y(0) = A = 0

y’(0) = Aλ + B = v_0

B = v_0

y(t) = v_0*t*e^(λt)

Overdamped:

λ = (-γ ± sqrt(γ^2-4mk))/(2m)

λ_1 = (-γ - sqrt(γ^2-4mk))/(2m)

λ_2 = (-γ + sqrt(γ^2-4mk))/(2m)

y_1 = e^(λ_1*t), y_2 = e^(λ_2*t)

y_g = Ae^(λ_1*t) + Be^(λ_2*t)

y_g’ = Aλ_1e^(λ_1*t) + Bλ_2e^(λ_2*t)

y(0) = 0 = A + B

A = -B

y’(0) = -Bλ_1 + Bλ_2 = v_0

B = -v_0/(λ_1-λ_2)

A = v_0/(λ_1-λ_2)

y(t) = (v_0/(λ_1-λ_2))*e^(λ_1*t) + (-v_0/(λ_1-λ_2))*e^(λ_2*t)

Underdamped:

λ = (-γ±i*sqrt(4mk-γ^2))/(2m) = (-γ/(2m)) ± i*(sqrt(4mk-γ^2)/(2m))

λ_1 = (-γ/(2m)) - i*(sqrt(4mk-γ^2)/(2m))

λ_2 = (-γ/(2m)) + i*(sqrt(4mk-γ^2)/(2m))

y_g = e^((-γ/(2m))t)(Acos((sqrt(4mk-γ^2)/(2m))*t) + Bsin((sqrt(4mk-γ^2)/(2m))*t))

y(0) = 0 = Acos(0)

A = 0

y’(0) = v_0 = B*y

B = v_0/y

y(t) = e^((-γ/(2m))t)((v_0/y)sin((sqrt(4mk-γ^2)/(2m))*t))

confidence rating #$&*:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK"

&#Good responses. See my notes and let me know if you have questions. &#