query_asst_9

course Mth 164

Where do the questions for this come from? Some are the same as on the paper problems but the majority I cannot find in the book or paper problems.

}??}??c·???G??P??·assignment #009009.

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Precalculus II

04-05-2007

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17:18:59

Query problem 7.3.14 b = 4, c = 1, `alpha = 120 deg

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RESPONSE -->

confidence assessment:

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17:19:32

specify the unknown sides and angles of your triangle.

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RESPONSE -->

a is approx. 4.583

'lamda is approx. 10.892

'beta is approx. 49.100

confidence assessment: 3

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17:21:11

Explain how you obtained your results.

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RESPONSE -->

I first found the value of a using the law of cosines (a^2=1^2+4^2-2(1*4)cos 120 deg). Then, using the found value for a (4.583) I set up the law of sines to solve for 'beta (sin120/4.583 = sin 'beta/4) and again to solve for 'lamda (sin120/4.583 = sin 'lamda/1).

confidence assessment: 3

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17:22:13

Query problem 7.3.32 500 ft tower 15 deg slope 2 guy wires 100 ft on either side of base on slope

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RESPONSE -->

This problem was not in Sullivan 7th edition, rather a 500 ft tower, a 10 deg slope at one side, flat on the other. The guy wire on the flat side is attatched half-way up the tower and the sloped side goes to the top, both extended outward 100 ft from the base.

confidence assessment:

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17:22:33

how long should the two guy wires be?

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The sloped side is 492.58ft and the flat side is 269.26ft.

confidence assessment: 3

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17:25:17

Explain how you obtained your result.

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RESPONSE -->

Fot both equations I used the law of cosines. On the slope side I set up a^2 = 500^2 + 100^2 - 2(500*100)cos 80deg because the tower is 500 ft tall, the guy wire is 100ft from the building's base and the angle of the base to ground instead of being 90 deg is 80 deg due to the 10 deg slope upwards.

The flat side wire length is found by using the equation b^2 = 250^2 + 100^2 - 2(250*100) cos 90 deg using the height at which the wire is attatched to the building, it is attatched half-way up a 500 ft building so 250 ft the wire is 100 ft from the base and the base to the ground angle is 90 deg since the ground is flat.

confidence assessment: 3

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17:25:23

query problem 7.5.6 coiled spring 10 cm displ period 3 sec moving downward at equil when t = 0.

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confidence assessment:

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17:25:48

What is the equation of motion of the object?

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RESPONSE -->

d = -10 cos (2'pi/3 t)

confidence assessment: 2

** Good. See below for two alternatives:

Modeling with reference-circle point moving counterclockwise at angular velocity omega on a circle of radius A, and using the sine function, displacement will be

y(t) = A sin (`omega * t + theta0), where theta0 is the initial angular position on the reference circle.

We know that the spring is at equilibrium when t = 0, so theta0 is either 0 or pi.

Since the spring is initially moving downward, we conclude that theta0 is pi.

The amplitude of motion will be 10 cm so A = 10 cm.

The period of motion is 2`pi/`omega = 3 sec, so `omega = 2`pi/(3 sec), so

the equation of motion is y = 10 cm sin( 2 pi / (3 sec) * t + pi).

Alternatively we could model using the cosine function, though this is less intuitive for an object moving up and down:

x(t) = A cos(omega * t + theta0),

in which case since the spring is at equilibrium when t = 0, theta0 is pi/2 (motion at t = 0 therefore being in the negative direction)

and we get

x(t) = 10 cm cos(2 pi / (3 sec) * t + pi/2).

**

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17:26:19

What is the amplitude of the motion and how does this amplitude affect the equation of motion?

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The amplitude of the motion is 10, and this affects the equation of the motion because it is 10 times the amount of the original equation of motion.

confidence assessment: 3

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17:26:52

What is the period of the motion, how does the period affect the angular frequence of the motion and how does the angular frequency appear in the equation of motion?

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RESPONSE -->

The period of motion is 3 sec, the time required to complete one vibration. Angular frequency is the reciprocal of the period, 1/3 oscillations/sec.

confidence assessment: 3

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17:27:44

What give condition determines the phase shift for this situation and how does the phase shift affect the equation of motion?

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The fact that at t=0 which is (0,0) on the graph the spring is moving downward, the cosine function is used so no phase shift will be required if this function is used.

confidence assessment: 2

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17:27:48

query problem 7.5.12 d=5 cos(`pi /2)t

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confidence assessment:

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17:28:58

Describe the motion of the object

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The object starts at a 5 unit displacement in the positive direction, when released will travel ina negative direction, past it's equilibrium and back in the positive direction losing some of it's momentum and coming to rest eventually at the point of equilibrium when t=0.

confidence assessment: 3

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17:29:08

What is the maximum displacement for its resting position?

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5

confidence assessment: 3

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17:30:02

What is the time required for the oscillation?

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period = 2'pi/'omega = 2'pi/('pi/2) = 2pi/1 * 2/'pi = 4'pi/'pi = 4 sec.

confidence assessment: 3

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17:30:25

What is the frequency?

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Frequency = 1/4 oscillations/sec since it is the reciprocal of the period, 4 sec.

confidence assessment: 3

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17:31:17

what is the area of the cone?

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This question 12 in sect. 7.5 is not the same in Sullivan 7th edition and is not one of the paper problems. I've been answering these questions from the information given and I do not know what this question means or pertains to.

confidence assessment:

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17:31:27

Explain how you obtained eac result

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As explained previously.

confidence assessment:

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17:31:29

Comm on any surprises or insights you experienced as a result of this assignment.

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confidence assessment:

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Your work on this assignment is very good. Let me know if you have questions.