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Homework 13 Solutions - homework 13 ALIBHAI ZAHID Due 4:00...

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homework 13 – ALIBHAI, ZAHID – Due: Apr 25 2007, 4:00 am 1 Question 1 part 1 of 3 10 points A 2 . 2 kg object oscillates with an initial am- plitude of 101 cm on a spring of force constant 98 . 8 N / m. Find the period. Correct answer: 0 . 93759 s (tolerance ± 1 %). Explanation: Let : m = 2 . 2 kg and k = 98 . 8 N / m . The period of the oscillator is T = 2 π radicalbigg m k = 2 π radicalBigg 2 . 2 kg 98 . 8 N / m = 0 . 93759 s . Question 2 part 2 of 3 10 points Find the total initial energy. Correct answer: 50 . 3929 J (tolerance ± 1 %). Explanation: Let : A = 101 cm = 1 . 01 m . The initial energy is E 0 = 1 2 k A 2 = 1 2 (98 . 8 N / m)(1 . 01 m) 2 = 50 . 3929 J . Question 3 part 3 of 3 10 points If the energy decreases by 1 percent per period, find the damping constant b . Correct answer: 0 . 0234644 kg / s (tolerance ± 1 %). Explanation: Let : | Δ E | E = 0 . 01 , m = 2 . 2 kg , and T = 0 . 93759 s . Relating the fractional rate at which the en- ergy decreases, Q = 2 π parenleftbigg | Δ E | E parenrightbigg cycle = 2 π 0 . 01 . Express the Q value in terms of b : Q = ω 0 m b = 2 π 0 . 01 b = 0 . 01 ω 0 m 2 π = 0 . 01 (2 π ) m 2 π T = 0 . 01 m T = 0 . 01 (2 . 2 kg) 0 . 93759 s = 0 . 0234644 kg / s . Question 4 part 1 of 1 10 points What beat frequencies are possible with tuning forks of frequencies 256, 259, and 261 Hz? 1. There are 3 possible beat frequencies, 515 Hz, 520 Hz, and 517 Hz. 2. There are 3 possible beat frequencies, 256 Hz, 259 Hz, and 261 Hz. 3. There are 4 possible beat frequencies, 2 Hz, 3 Hz, 5 Hz, and 7 Hz. 4. There are 3 possible beat frequencies, 2 Hz, 3 Hz, and 5 Hz. correct
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homework 13 – ALIBHAI, ZAHID – Due: Apr 25 2007, 4:00 am 2 5. There are 2 possible beat frequencies, 2 Hz and 3 Hz. Explanation: The beat frequencies are the differences in the fork frequencies: 261 Hz - 259 Hz = 2 Hz , 261 Hz - 256 Hz = 5 Hz , 259 Hz - 256 Hz = 3 Hz . Question 5 part 1 of 1 10 points A wave pulse on a string is described by the equation y 1 = A ( B x - C t ) 2 + D . A second wave pulse on the same string is described by y 2 = - A ( B x + C t - E ) 2 + D , where x is in meters and t in seconds, and A = 5 . 7 m, B = 3 . 58 m 1 , C = 6 s 1 , D = 4 . 59, and E = 7 . 08. At what time will the two waves exactly cancel everywhere? Correct answer: 0 . 59 s (tolerance ± 1 %). Explanation: Basic Concepts: Superposition principle: y = y 1 ( x, t ) + y 2 ( x, t ) Solution: Since y = y 1 + y 2 , y will be zero when y 1 + y 2 = 0 . This implies, after some algebra, that B x - C t = ± ( B x + C t - E ) (3 . 58 m 1 ) x - (6 s 1 ) t = ± [(3 . 58 m 1 ) x + (6 s 1 ) t - 7 . 08] . (The ± arises from taking a square root.) When solving the “+” equation, we get 2 C t = E , therefore t = E 2 C = (7 . 08) 2 (6 s 1 ) = 0 . 59 s . Because x cancels out, our solution tells us y is zero for all positions x at this particular time t . Question 6 part 1 of 1 10 points Two wires are made of the same material but the second wire has twice the diameter and twice the length of the first wire. When the two wires are stretched, and the tension in the second wire is also twice the tension in the first wire, the fundamental frequency of the first wire is 920 Hz.
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