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hw12 - al-ameri(aoa434 Hwk12 Stokes(19102 This print-out...

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al-ameri (aoa434) – Hwk12 – Stokes – (19102) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is ONLINE HOMEWORK No 12. It is due before 6.00 am on Tuesday 28th April, Abu Dhabi Time (9.00 pm Monday in Texas) 001 10.0 points The equation of motion of a simple harmonic oscillator is d 2 x dt 2 = 9 x , where x is displacement and t is time. What is the period of oscillation? 1. T = 2 π 9 2. T = 3 2 π 3. T = 9 2 π 4. T = 2 π 3 correct 5. T = 6 π Explanation: d 2 x dt 2 = ω 2 x , where ω is the angular frequency, so the pe- riod of oscillation is T = 2 π ω = 2 π 9 = 2 π 3 . 002 (part 1 of 3) 10.0 points A 668 g mass is connected to a light spring of force constant 2 N / m that is free to oscillate on a horizontal, frictionless track. The mass is displaced 8 cm from the equilibrium point and released from rest. 2 N / m 668 g 8 cm x = 0 x Find the period of the motion. Correct answer: 3 . 63122 s. Explanation: Let : m = 668 g = 0 . 668 kg and k = 2 N / m . This situation corresponds to the special case x ( t ) = A cos ωt , so the frequency is ω = radicalbigg k m = radicalBigg 2 N / m 0 . 668 kg = 1 . 73032 s 1 and the period is T = 2 π ω = 2 π 1 . 73032 s 1 = 3 . 63122 s . 003 (part 2 of 3) 10.0 points What is the maximum speed of the mass? Correct answer: 0 . 138426 m / s. Explanation: Let : A = 8 cm . The velocity as a function of time is v ( t ) = ω A sin( ω t ) , so the maximum speed of the mass is v max = ω A = (1 . 73032 s 1 ) (0 . 08 m) = 0 . 138426 m / s . 004 (part 3 of 3) 10.0 points What is the maximum acceleration of the mass? Correct answer: 0 . 239521 m / s 2 .
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al-ameri (aoa434) – Hwk12 – Stokes – (19102) 2 Explanation: The acceleration as a function of time is a ( t ) = ω 2 A cos( ω t ) , so the maximum acceleration of the mass is a max = ω 2 A = (1 . 73032 s 1 ) 2 (0 . 08 m) = 0 . 239521 m / s 2 . 005 (part 1 of 4) 10.0 points A 15 . 1 kg mass is suspended on a 1 × 10 5 N / m spring. The mass oscillates up and down from the equilibrium position y eq = 0 according to y ( t ) = A sin( ωt + φ 0 ) . Find the angular frequency of the oscillat- ing mass. Correct answer: 81 . 3788 s 1 . Explanation: Let : M = 15 . 1 kg and k = 1 × 10 5 N / m . When the mass moves out of equilibrium, it suffers a net restoring force F net y = F spring Mg = k ( y y eq ) = ky , and accelerates back towards the equilibrium position at the rate a y = F net y M = k M y . Therefore, the mass oscillates harmonically with angular frequency ω = radicalbigg a y y = radicalbigg k M = radicalBigg 1 × 10 5 N / m 15 . 1 kg = 81 . 3788 s 1 . 006 (part 2 of 4) 10.0 points At time t 0 = 0 the mass is at 19 . 2 cm and moving upward at velocity +21 . 1 m / s. Find the amplitude of the oscillating mass. Correct answer: 32 . 2631 cm. Explanation: Let : y 0 = 19 . 2 cm and v 0 = 21 . 1 m / s . The mass oscillates according to the SHM equation y ( t ) = A sin( ωt + φ 0 ) , so its velocity is v y ( t ) = d y dt = A ω cos( ωt + φ 0 ) . At time t 0 = 0, we have y 0 y ( t = 0) = A sin φ 0 and v 0 = v y ( t = 0) = A ω cos φ 0 , so A sin φ 0 = y 0 and A cos φ 0 = v 0 ω . (1) Consequently, A 2 = ( A sin φ 0 ) 2 + ( A cos φ 0 ) 2 = ( y 0 ) 2 + parenleftBig v 0 ω parenrightBig 2 A = radicalbigg ( y 0 ) 2 + parenleftBig v 0 ω parenrightBig 2 = radicalBigg (19 . 2 cm) 2 + parenleftbigg 21 . 1 m / s 81 . 3788 s 1 parenrightbigg 2 = 32 . 2631 cm .
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