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Unformatted text preview: Su (ycs73) – HW10 – Tsoi – (58020) 1 This printout should have 15 questions. Multiplechoice 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( ωt + φ ) , with amplitude 5 m, angular frequency π s − 1 , and initial phase π 3 radians. Every now and then, the particle’s 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 , ω = π s − 1 , and φ = π 3 . x = A cos( ω t + φ ) v = − Aπ sin( ω t + φ ) and ω = radicalbigg k m . K = U 1 2 mv 2 = 1 2 k x 2 A 2 ω 2 sin 2 ( ω t + φ ) = k m A 2 cos 2 ( ω t + φ ) ω radicalBig sin 2 ( ω t + φ ) = radicalbigg k m radicalBig cos 2 ( ω t + φ ) vextendsingle vextendsingle vextendsingle sin parenleftBig π t + π 3 parenrightBigvextendsingle vextendsingle vextendsingle = vextendsingle vextendsingle vextendsingle cos parenleftBig π t + π 3 parenrightBigvextendsingle vextendsingle vextendsingle . Since  sin x  =  cos x  when x = (2 n + 1) π 4 , n = 0 , ± 1 , ± 2 ,..., (2 n + 1) π 4 = π t + π 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 θ = θ ) 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 α 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 ◦ α 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 , ℓ = 11 . 4 m , d = 5 m , ℓ − d = 6 . 4 m , θ = 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 ℓ − 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 π radicalBigg ℓ g = 2 π radicalBigg (11 . 4 m) (9 . 8 m / s 2 ) = 6 . 77672 s , and T 2 = 2 π radicalBigg...
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This note was uploaded on 04/15/2009 for the course PHY 58020 taught by Professor Tsoi during the Spring '09 term at University of Texas.
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