hmwk10 - Su (ycs73) HW10 Tsoi (58020) 1 This print-out...

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Unformatted text preview: Su (ycs73) HW10 Tsoi (58020) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 10.0 points A particle oscillates harmonically x = A sin( t + ) , with amplitude 5 m, angular frequency s 1 , and initial phase 3 radians. Every now and then, the particles kinetic energy and poten- tial energy happen to be equal to each other ( K = U ). When does this equality happen for the first time after t = 0? 1. 0.2238 s 2. 0.4167 s correct 3. 0.9967 s 4. 0.8623 s 5. 0.3467 s 6. 0.6547 s 7. 0.5884 s 8. 0.5267 s 9. 0.1294 s 10. 0.7615 s Explanation: Let : A = 5 m , = s 1 , and = 3 . x = A cos( t + ) v = A sin( t + ) and = radicalbigg k m . K = U 1 2 mv 2 = 1 2 k x 2 A 2 2 sin 2 ( t + ) = k m A 2 cos 2 ( t + ) radicalBig sin 2 ( t + ) = radicalbigg k m radicalBig cos 2 ( t + ) vextendsingle vextendsingle vextendsingle sin parenleftBig t + 3 parenrightBigvextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle cos parenleftBig t + 3 parenrightBigvextendsingle vextendsingle vextendsingle . Since | sin x | = | cos x | when x = (2 n + 1) 4 , n = 0 , 1 , 2 ,..., (2 n + 1) 4 = t + 3 2 n + 1 4 = t + 1 3 t = 2 n + 1 4 1 3 . Since t > 0 for n 1, the first positive time is when t = 1: t = 0 . 416667 002 10.0 points Note: The pendulum bob is released at a height below the height of the peg. Assume: Use the small angle approximate (sin = ) to calculate the period. A pendulum made of a string of length 11 . 4 m and a spherical bob of mass 1 . 5 kg is able to swing in a vertical plane. The pen- dulum is released from an angular position 53 from vertical as shown in the figure be- low. The string hits a peg located a distance 5 m below the point of suspension and swings about the peg up to an angle on the other side of the peg. Then, the bob proceeds to oscillate back and forth between these two angular extremities. The acceleration of gravity is 9 . 8 m / s 2 . Su (ycs73) HW10 Tsoi (58020) 2 9 . 8 m / s 2 1 . 5 kg 1 1 . 4 m 53 5m What is the period of the pendulum plus peg system as shown above? Correct answer: 5 . 92715 s. Explanation: Let : m = 1 . 5 kg , = 11 . 4 m , d = 5 m , d = 6 . 4 m , = 53 , and g = 9 . 8 m / s 2 . As shown in the figure, the motion can be broken into two parts: that before the string contacts the peg and that after it makes con- tact. Its motion on the right half of the diagram is that of a pendulum with length 11 . 4 m and period T 1 ; its motion on the left side acts as a pendulum with length d and period T 2 . For small angles from the verti- cal, these periods are given by the familiar equation for a simple pendulum, T 1 = 2 radicalBigg g = 2 radicalBigg (11 . 4 m) (9 . 8 m / s 2 ) = 6 . 77672 s , and T 2 = 2 radicalBigg...
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hmwk10 - Su (ycs73) HW10 Tsoi (58020) 1 This print-out...

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