old33 - hernandez (ejh742) oldhomework 33 Turner (58120) 1...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hernandez (ejh742) oldhomework 33 Turner (58120) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points Two waves in one string are described by the relationships y 1 = A 1 cos( k 1 x- 1 t ) y 2 = A 2 sin( k 2 x- 2 t ) where A 1 = 4 . 2 cm, A 2 = 4 . 9 cm, k 1 = 6 cm 1 , k 2 = 5 cm 1 , 1 = 3 rad / s, 2 = 1 rad / s, y and x are in centimeters, and t is in seconds. Find the superposition of the waves at the position 1 cm and time 1 s. Correct answer:- 7 . 8663 cm. Explanation: Let : x 1 = 1 cm and t 1 = 1 s . At this point we have y 1 = (4 . 2 cm) cos bracketleftBig (6 cm 1 ) (1 cm)- (3 rad / s) (1 s) bracketrightBig =- 4 . 15797 cm and y 2 = (4 . 9 cm) sin bracketleftBig (5 cm 1 ) (1 cm)- (1 rad / s) (1 s) bracketrightBig =- 3 . 70833 cm , so y 1 + y 2 =- 7 . 8663 cm . 002 (part 2 of 3) 10.0 points Find the superposition of the waves y 1 + y 2 at the position 0 . 6 cm and time 0 . 7 s. Correct answer: 3 . 95105 cm. Explanation: Let : x 2 = 0 . 6 cm and t 2 = 0 . 7 s . At this point we have y 1 = (4 . 2 cm) cos bracketleftBig (6 cm 1 ) (0 . 6 cm)- (3 rad / s) (0 . 7 s) bracketrightBig = 0 . 297096 cm and y 2 = (4 . 9 cm) sin bracketleftBig (5 cm 1 ) (0 . 6 cm)- (1 rad / s) (0 . 7 s) bracketrightBig = 3 . 65396 cm , so y 1 + y 2 = 3 . 95105 cm . 003 (part 3 of 3) 10.0 points Find the superposition of the waves y 1 + y 2 at the position 0 . 4 cm and time- 28 s. Correct answer:- 4 . 81531 cm. Explanation: Let : x 3 = 0 . 4 cm and t 3 =- 28 s . At this point we have y 1 = (4 . 2 cm) cos bracketleftBig (6 cm 1 ) (0 . 4 cm)- (3 rad / s) (- 28 s) bracketrightBig = 0 . 0260483 cm and y 2 = (4 . 9 cm) sin bracketleftBig (5 cm 1 ) (0 . 4 cm)- (1 rad / s) (- 28 s) bracketrightBig =- 4 . 84135 cm , so y 1 + y 2 =- 4 . 81531 cm . 004 10.0 points The distance between two successive maxima of a certain transverse wave is 1 . 22 m. Eight hernandez (ejh742) oldhomework 33 Turner (58120) 2 crests, or maxima, pass a given point along the direction of travel every 13 . 2 s. Calculate the wave speed. Correct answer: 0 . 739394 m / s. Explanation: Let : = 1 . 22 m , t = 13 . 2 s , and n = 8 . The frequency of the transverse wave is f = n t = 8 13 . 2 s = 0 . 606061 Hz . The wave speed then is v = f = (1 . 22 m)(0 . 606061 Hz) = . 739394 m / s . 005 (part 1 of 3) 10.0 points The time needed for a water wave to change from the equilibrium level to the crest is . 5741 s....
View Full Document

Page1 / 7

old33 - hernandez (ejh742) oldhomework 33 Turner (58120) 1...

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