Homework 13 Solutions - homework 13 GADHIA, TEJAS Due: Apr...

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Unformatted text preview: homework 13 GADHIA, TEJAS Due: Apr 25 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 . 8 cm, A 2 = 4 . 1 cm, k 1 = 3 cm 1 , k 2 = 5 cm 1 , 1 = 4 rad / s, 2 = 1 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 . 9 cm and time t 1 = 2 s. Correct answer: 4 . 56036 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 8 cm) cos bracketleftBig (3 cm 1 ) (0 . 9 cm) (4 rad / s) (2 s) bracketrightBig = 2 . 10662 cm y 2 = (4 . 1 cm) sin bracketleftBig (5 cm 1 ) (0 . 9 cm) (1 rad / s) (2 s) bracketrightBig = 2 . 45374 cm , so y 1 + y 2 = 4 . 56036 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: . 0320272 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 8 cm) cos bracketleftBig (3 cm 1 ) (1 cm) (4 rad / s) (0 . 7 s) bracketrightBig = 3 . 72425 cm y 2 = (4 . 1 cm) sin bracketleftBig (5 cm 1 ) (1 cm) (1 rad / s) (0 . 7 s) bracketrightBig = 3 . 75628 cm , so y 1 + y 2 = . 0320272 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 . 3 cm and time t 3 = 91 s. Correct answer: . 657081 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 8 cm) cos bracketleftBig (3 cm 1 ) (0 . 3 cm) (4 rad / s) ( 91 s) bracketrightBig = 3 . 37887 cm y 2 = (4 . 1 cm) sin bracketleftBig (5 cm 1 ) (0 . 3 cm) (1 rad / s) ( 91 s) bracketrightBig = 4 . 03595 cm , so y 1 + y 2 = . 657081 cm . Question 4 part 1 of 1 0 points An incident sine wave y i ( x, t ) = A sin( k x t ) (1) travels to the right and is reflected off a fixed end of the string at x = L . Which of the following expression gives the correct formula for the reflected wave y r ( x, t )? 1. y r ( x, t ) = A sin( k x + t 2 k L ) cor- rect homework 13 GADHIA, TEJAS Due: Apr 25 2007, 4:00 am 2 2. y r ( x, t ) = A sin( k x + t + k L ) 3. y r ( x, t ) = A sin( k x + t + 2 k L ) 4. y r ( x, t ) = A sin( k x + t ) 5. y r ( x, t ) = A sin( k x t + 2 k L ) 6. y r ( x, t ) = A sin( k x + t k L ) 7. y r ( x, t ) = A sin( k x t ) 8. y r ( x, t ) = A sin( k x t + k L ) 9. y r ( x, t ) = A sin( k x t k L ) 10. y r ( x, t ) = A sin( k x t 2 k L ) Explanation: When a wave is reflected at point x = L , it continues to travel in the reverse direction from this point thus ( x i L ) mapsto ( x r L ) or in other words, x i mapsto x r = 2 L x i . (2) Also, when the wave is reflected off a fixed end of the string, the transverse displacement y changes sign, thus y r ( x r , t ) = y i ( x i , t ) . (3) Combining eqs. (1) and (2) together, we arriveCombining eqs....
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Homework 13 Solutions - homework 13 GADHIA, TEJAS Due: Apr...

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