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Unformatted text preview: practice 13 ALIBHAI, ZAHID Due: Apr 22 2007, 4:00 am 1 Question 1 part 1 of 3 10 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 = 2 . 2 cm, A 2 = 2 . 7 cm, k 1 = 6 cm 1 , k 2 = 4 cm 1 , 1 = 3 rad / s, 2 = 4 rad / s, y and x are in centimeters, and t is in seconds. Find the superposition of the waves y 1 + y 2 at the position x 1 = 0 . 5 cm and time t 1 = 2 s. Correct answer: 1 . 42356 cm (tolerance 1 %). Explanation: At this point we have y 1 = (2 . 2 cm) cos bracketleftBig (6 cm 1 ) (0 . 5 cm) (3 rad / s) (2 s) bracketrightBig = 2 . 17798 cm y 2 = (2 . 7 cm) sin bracketleftBig (4 cm 1 ) (0 . 5 cm) (4 rad / s) (2 s) bracketrightBig = 0 . 754422 cm , so y 1 + y 2 = 1 . 42356 cm . Question 2 part 2 of 3 10 points Find the superposition of the waves y 1 + y 2 at the position x 2 = 1 cm and time t 2 = 0 . 7 s. Correct answer: 0 . 919455 cm (tolerance 1 %). Explanation: At this point we have y 1 = (2 . 2 cm) cos bracketleftBig (6 cm 1 ) (1 cm) (3 rad / s) (0 . 7 s) bracketrightBig = 1 . 59705 cm y 2 = (2 . 7 cm) sin bracketleftBig (4 cm 1 ) (1 cm) (4 rad / s) (0 . 7 s) bracketrightBig = 2 . 51651 cm , so y 1 + y 2 = 0 . 919455 cm . Question 3 part 3 of 3 10 points Find the superposition of the waves y 1 + y 2 at the position x 3 = 0 . 5 cm and time t 3 = . 38 s. Correct answer: 2 . 1891 cm (tolerance 1 %). Explanation: At this point we have y 1 = (2 . 2 cm) cos bracketleftBig (6 cm 1 ) (0 . 5 cm) (3 rad / s) ( . 38 s) bracketrightBig = 1 . 19161 cm y 2 = (2 . 7 cm) sin bracketleftBig (4 cm 1 ) (0 . 5 cm) (4 rad / s) ( . 38 s) bracketrightBig = . 997491 cm , so y 1 + y 2 = 2 . 1891 cm . Question 4 part 1 of 2 10 points Two sound sources radiating in phase at a frequency of 540 Hz interfere such that max ima are heard at angles of 0 and 24 from a line perpendicular to that joining the two sources. The velocity of sound is 340 m/s. practice 13 ALIBHAI, ZAHID Due: Apr 22 2007, 4:00 am 2 y L d S 1 S 2 listening direction Find the separation between the two sources. Correct answer: 1 . 548 m (tolerance 1 %). Explanation: Let : f = 540 Hz , v = 340 m / s , 1 = 0 , and 2 = 24 . r 2 r 1 y L d S 1 S 2 = ta n 1 parenleftBig y L parenrightBig listening direction d sin r 2 r 1 P O negationslash S 2 Q S 1 90 Q 2 or 360 = 500 1 1 90 180 270 360 / 2 3 / 2 2 2 = = d sin = d y radicalbig L 2 + y 2 d y L Because a maximum is heard at 0 and the sources are in phase, we can conclude that the path difference is 0. Because the next maximum is heard at 24 , the path difference to that position must be one wavelength: sin = s d d = sin = v f sin = 340 m / s (540 Hz) sin24 = 1 . 548 m...
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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.
 Spring '08
 Turner
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