Homework 11 Sol. - Abd Elhai, Mohamed – Homework 11 –...

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Unformatted text preview: Abd Elhai, Mohamed – Homework 11 – Due: Apr 30 2008, 9:00 pm – Inst: Dr Kevin Dean 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. This is ONLINE HOMEWORK No. 11 It is due by 06:00 AM Abu Dhabi Time on Tuesday 29 April (9:00 PM Monday 28 April, Texas time). 001 (part 1 of 2) 4 points A piston in an automobile engine is in simple harmonic motion. Its amplitude of oscillation from the equilibrium (centered) position is ± 2 . 5 cm and its mass is 3 . 15 kg. Find the maximum velocity of the piston when the auto engine is running at the rate of 2600 rev / min. Correct answer: 6 . 80678 m / s. Explanation: Let : A = 2 . 5 cm , m = 3 . 15 kg , and f = 2600 rev / min . ω = 2 π f = 2 π (2600 rev / min) 1 min 60 s = 272 . 271 rad / s . The simple harmonic motion is described by x = A cos ω t, where ω is the frequency in rad/s if t is in seconds. The velocity is v = dx dt =- Aω sin( ω t ) meaning the maximum velocity is v max = Aω = (0 . 025 m) (272 . 271 rad / s) = 6 . 80678 m / s , since sine has a maximum value of 1. 002 (part 2 of 2) 4 points Find the maximum acceleration of the piston when the auto engine is running at this rate. Correct answer: 1 . 85329 km / s 2 . Explanation: The acceleration is a = d 2 x dt 2 = dv dt =- Aω 2 cos( ω t ) , so the maximum acceleration is a max = Aω 2 = (0 . 025 m) (272 . 271 rad / s) 2 = 1 . 85329 km / s 2 , since cosine has a maximum value of 1. keywords: 003 (part 1 of 1) 4 points Simple harmonic motion can be described us- ing the equation y = A sin( k x- ω t- φ ) . Consider the simple harmonic motion given by the figure. + A-A y π 2 π 3 π 4 π At position x = 0, we have ω t What equation describes the motion? sin(- θ ) =- sin θ . 1. y = A sin ‡- ω t + π 2 · 2. y = A sin µ- ω t- 3 π 2 ¶ Abd Elhai, Mohamed – Homework 11 – Due: Apr 30 2008, 9:00 pm – Inst: Dr Kevin Dean 2 3. y = A cos µ- ω t- 3 π 2 ¶ 4. y = A cos ‡- ω t + π 2 · 5. y = A sin ‡- ω t- π 2 · correct 6. y = A tan ‡- ω t + π 2 · 7. y = A cos ‡- ω t- π 2 · 8. y = A tan µ- ω t- 3 π 2 ¶ 9. y = A tan ‡- ω t- π 2 · Explanation: If the y-axis is moved such that the angle- ω t- π 2 = 0 , the curve is the sine function A sin(- ω t ) ; therefore y = A sin ‡- ω t- π 2 · is the curve in the figure. keywords: 004 (part 1 of 4) 3 points A 13 . 8 kg mass is suspended on a 100000 N / m spring. The mass oscillates up and down from the equilibrium position y eq = 0 according to y ( t ) = A sin( ωt + φ ) . Find the angular frequency of the oscillat- ing mass. Correct answer: 85 . 1257 s- 1 ....
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Homework 11 Sol. - Abd Elhai, Mohamed – Homework 11 –...

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