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Unformatted text preview: homework 12 – VENNES, ROSS – Due: Nov 21 2007, 1:00 am 1 Question 1, chap 16, sect 2. part 1 of 1 10 points What is the frequency corresponding to a period of 0 . 01 s? Correct answer: 100 Hz (tolerance ± 1 %). Explanation: Let : t = 0 . 01 s . f = 1 t = 1 . 01 s = 100 Hz . Question 2, chap 16, sect 2. part 1 of 1 10 points A harmonic wave is traveling along a rope. The oscillator that generates the wave com pletes 41.0 vibrations in 23.1 s. A given crest of the wave travels 383 cm along the rope in a time period of 13.5 s. What is the wavelength? Correct answer: 0 . 159843 m (tolerance ± 1 %). Explanation: Basic Concepts: v = f λ f = number of vibrations t Δ x = v Δ t Given: n = 41 . 0 vibrations t = 23 . 1 s Δ x = 383 cm Δ t = 13 . 5 s Solution: λ = v f = Δ x Δ t n t = Δ xt n Δ t = (383 cm)(23 . 1 s) 41 (13 . 5 s) · 1 m 100 cm = 0 . 159843 m Question 3, chap 16, sect 2. part 1 of 2 10 points Tension is maintained in a string as in the figure. The observed wave speed is 20 m / s when the suspended mass is 2 . 5 kg . The acceleration of gravity is 9 . 8 m / s 2 . 2 . 5 kg What is the mass per unit length of the string? Correct answer: 0 . 06125 kg / m (tolerance ± 1 %). Explanation: Let : v = 20 m / s and m = 2 . 5 kg . The tension in the string is F = mg and its velocity is v = radicalBigg F μ , and μ = F v 2 = mg v 2 = (2 . 5 kg) ( 9 . 8 m / s 2 ) (20 m / s) 2 = . 06125 kg / m . Question 4, chap 16, sect 2. part 2 of 2 10 points homework 12 – VENNES, ROSS – Due: Nov 21 2007, 1:00 am 2 What is the wave speed when the suspended mass is 2 kg? Correct answer: 17 . 8885 m / s (tolerance ± 1 %). Explanation: Let : m = 2 kg . The velocity is v = radicalBigg F μ = radicalbigg mg μ = radicalBigg (2 kg) (9 . 8 m / s 2 ) . 06125 kg / m = 17 . 8885 m / s . Question 5, chap 16, sect 2. part 1 of 1 10 points A light string of mass 10 . 4 g and length 3 . 19 m has its ends tied to two walls that are separated by the distance 2 . 33 m. Two objects, each of mass 1 . 23 kg, are suspended from the string as in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 1 . 23 kg 1 . 23 kg . 7 9 7 5 m 1 . 595 m . 7 9 7 5 m 2 . 33 m A B θ If a wave pulse is sent from point A , how long does it take to travel to point B ? Correct answer: 36 . 4029 ms (tolerance ± 1 %). Explanation: Given : m = 10 . 4 g = 0 . 0104 kg , M = 1 . 23 kg , L = 3 . 19 m , L 2 = 1 . 595 m , L 4 = 0 . 7975 m , and D = 2 . 33 m . M M L 4 L 2 L 4 D A B θ θ d d Observing the lengths, 2 d + L 2 = D 4 d + L = 2 D d = 2 D L 4 and cos θ = d L 4 = 4 d L = 2 D L L , so θ = cos − 1 bracketleftbigg 2 D L 1 bracketrightbigg = cos − 1 bracketleftbigg 2 (2 . 33 m) 3 . 19 m 1 bracketrightbigg = 62 . 5603 ◦ ....
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This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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