Homework 13 Solutions - homework 13 ALIBHAI, ZAHID Due: Apr...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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 r m k = 2 π R 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, ±nd 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 p | Δ E | E P 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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 diFerences 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
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 9

Homework 13 Solutions - homework 13 ALIBHAI, ZAHID Due: Apr...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online