Solve Q4 please 3. (Free-driving force, damped mass-spring system). Recall that the underdamped motion of the systemis give by the IVPy" + 2xy +wy = 0,y(0) = yo, y'(0) = 20where A is the damping ratio, wo is the natural frequency of the system, yo is the initial positionof the mass, and vo is the initial velocity of the mass. The system motion (i.e., general solutionof the system) is given byy(t) = e-wostwhere = Show that[c₁ cos (wo√1 - 5²t) + ₂ sin(wo√/1-²t)]Swoyo +00Wo 1-²You may refer to the case of undamped motion studied in the class.01 = 900, C2 = 4. A-kg object stretches a spring with stiffness 18 N/m and damping coefficient kg/sec. There isno external driving force acting on the system. The spring initially displaced m upward from itsequilibrium position and given an initial velocity 1 m/sec. The oscillator DE and initial conditionsdescribing the system motion is give15+ y + 18y = 0,y (0)y'(0) = 1a. Use the relations in Question 3 to find the displacement (position) of the mass at any time.b. Use the relations/information from Question 3 and Part a. to find the displacement of themass at any time using the alternative formulay(t) = Re-wost cos(wo √1-(²t - 6)where R= √+ and the phase angle = tan-¹.

y"+2λy'+ω02y=0, y0=y0, y'0=v0yt=e-ω0ζtc1cosω01-ζ2t+c2sinω01-ζ2tζ=λω0c1=y0c2=ζω0y0+v0ω01-ζ2 Given…

Answered: 4. A-kg object stretches a spring with…

Trigonometry (MindTap Course List)

10th Edition

ISBN:9781337278461

Author:Ron Larson

Publisher:Ron Larson

Chapter3: Additional Topics In Trigonometry

Section3.3: Vectors In The Plane

Problem 11ECP

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Related questions

Question

Solve Q4 please

3. (Free-driving force, damped mass-spring system). Recall that the underdamped motion of the system
is give by the IVP
y" + 2xy +wy = 0,
y(0) = yo, y'(0) = 20
where A is the damping ratio, wo is the natural frequency of the system, yo is the initial position
of the mass, and vo is the initial velocity of the mass. The system motion (i.e., general solution
of the system) is given by
y(t) = e-wost
where = Show that
[c₁ cos (wo√1 - 5²t) + ₂ sin(wo√/1-²t)]
Swoyo +00
Wo 1-²
You may refer to the case of undamped motion studied in the class.
01 = 900, C2 =

4. A-kg object stretches a spring with stiffness 18 N/m and damping coefficient kg/sec. There is
no external driving force acting on the system. The spring initially displaced m upward from its
equilibrium position and given an initial velocity 1 m/sec. The oscillator DE and initial conditions
describing the system motion is give
1
5
+ y + 18y = 0,
y (0)
y'(0) = 1
a. Use the relations in Question 3 to find the displacement (position) of the mass at any time.
b. Use the relations/information from Question 3 and Part a. to find the displacement of the
mass at any time using the alternative formula
y(t) = Re-wost cos(wo √1-(²t - 6)
where R= √+ and the phase angle = tan-¹.

Expert Solution

Step 1:Find the displacement of the mass at any time

$y " + 2 λ y' + ω_{0}^{2} y = 0, y (0) = y_{0}, y' (0) = v_{0} y (t) = e^{- ω_{0} ζ t} [c_{1} \cos (ω_{0} \sqrt{1 - ζ^{2}} t) + c_{2} \sin (ω_{0} \sqrt{1 - ζ^{2}} t)] ζ = \frac{λ}{ω_{0}} c_{1} = y_{0} c_{2} = \frac{ζ ω_{0} y_{0} + v_{0}}{ω_{0} \sqrt{1 - ζ^{2}}}$

Given

$\frac{1}{2} y " + \frac{5}{2} y' + 18 y = 0, y (0) = - \frac{1}{2}, y' (0) = 1 \Rightarrow y " + 5 y' + 36 y = 0$

Now comparing we have

$∴ 2 λ = 5 \Rightarrow λ = \frac{5}{2} ω_{0}^{2} = 36 \Rightarrow ω_{0} = \pm 6 y_{0} = - \frac{1}{2} v_{0} = 1$

Considering $ω_{0} = 6$

$c_{1} = y_{0} = - \frac{1}{2} ζ = \frac{λ}{ω_{0}} = \frac{5}{2 \times 6} = \frac{5}{12} c_{2} = \frac{ζ ω_{0} y_{0} + v_{0}}{ω_{0} \sqrt{1 - ζ^{2}}} = \frac{\frac{5}{12} \times 6 \times (- \frac{1}{2}) + 1}{6 \sqrt{1 - {(\frac{5}{12})}^{2}}} = - \frac{1}{2 \sqrt{119}} 1 - ζ^{2} t = 1 - \frac{25 t}{144} ∴ y (t) = e^{- ω_{0} ζ t} [c_{1} \cos (ω_{0} \sqrt{1 - ζ^{2}} t) + c_{2} \sin (ω_{0} \sqrt{1 - ζ^{2}} t)] = e^{- 6 \cdot \frac{5}{12} \cdot t} [- \frac{1}{2} \cos (6 \sqrt{1 - \frac{25}{144}} t) - \frac{1}{2 \sqrt{119}} \sin (6 \sqrt{1 - \frac{25}{144}} y)] = e^{- \frac{5 t}{2}} [- \frac{1}{2} \cos (\frac{1}{2} \sqrt{119} t) - \frac{1}{2 \sqrt{119}} \sin (\frac{1}{2} \sqrt{119} t)]$

Hence the displacement at any time $t$ is $y (t) = e^{- \frac{5 t}{2}} [- \frac{1}{2} \cos (\frac{1}{2} \sqrt{119} t) - \frac{1}{2 \sqrt{119}} \sin (\frac{1}{2} \sqrt{119} t)]$