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Unformatted text preview: homework 14 FRENNEA, KYLE Due: May 1 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 = 3 . 5 cm, A 2 = 2 . 4 cm, k 1 = 4 cm 1 , k 2 = 4 cm 1 , 1 = 1 rad / s, 2 = 3 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 . 7 cm and time t 1 = 1 s. Correct answer: 1 . 27201 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 5 cm) cos bracketleftBig (4 cm 1 ) (0 . 7 cm) (1 rad / s) (1 s) bracketrightBig = . 795207 cm y 2 = (2 . 4 cm) sin bracketleftBig (4 cm 1 ) (0 . 7 cm) (3 rad / s) (1 s) bracketrightBig = . 476807 cm , so y 1 + y 2 = 1 . 27201 cm . Question 2 part 2 of 3 10 points Find the superposition of the waves y 1 + y 2 at the position x 2 = 2 cm and time t 2 = 0 . 8 s. Correct answer: 0 . 61419 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 5 cm) cos bracketleftBig (4 cm 1 ) (2 cm) (1 rad / s) (0 . 8 s) bracketrightBig = 2 . 12923 cm y 2 = (2 . 4 cm) sin bracketleftBig (4 cm 1 ) (2 cm) (3 rad / s) (0 . 8 s) bracketrightBig = 1 . 51504 cm , so y 1 + y 2 = 0 . 61419 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 = 60 s. Correct answer: 1 . 68905 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 5 cm) cos bracketleftBig (4 cm 1 ) (0 . 5 cm) (1 rad / s) (60 s) bracketrightBig = 0 . 41713 cm y 2 = (2 . 4 cm) sin bracketleftBig (4 cm 1 ) (0 . 5 cm) (3 rad / s) (60 s) bracketrightBig = 2 . 10618 cm , so y 1 + y 2 = 1 . 68905 cm . Question 4 part 1 of 2 10 points Two sound sources radiating in phase at a frequency of 460 Hz interfere such that max ima are heard at angles of 0 and 29 from a line perpendicular to that joining the two sources. The velocity of sound is 340 m/s. y L d S 1 S 2 listening direction homework 14 FRENNEA, KYLE Due: May 1 2007, 4:00 am 2 Find the separation between the two sources. Correct answer: 1 . 52458 m (tolerance 1 %). Explanation: Let : f = 460 Hz , v = 340 m / s , 1 = 0 , and 2 = 29 . 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 29 , the path difference to that position must be one wavelength: sin = s d d = sin = v f sin = 340 m / s (460 Hz) sin29 = 1 . 52458 m ....
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 Spring '08
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