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Navneet kumar
37 Points
9 years ago
Let the amplitude of Incident and Reflected Waves be A1 and A2. We know that Energy of wave is directly proportional to its square of amplitude. Given, A m p l i t u d e o f N o d e A m p l i t u d e o f A n t n o d e = 4 1 ? A 1 + A 2 A 1 - A 2 = 4 k 1 k where k is proportional constant. Solving above equation we get A 1 = 5 k / 2 a n d A 2 = 3 k / 2 So, Energy of incident and reflected wave is given as :- E i = C ( 5 k / 2 ) 2 a n d E r = C ( 3 k / 2 ) 2 where C is proportionality constant Hence Energy of transmitted wave will be E t = E i - E r = C ( 5 k / 2 ) 2 - C ( 3 k / 2 ) 2 = C ( 16 k 2 / 4 ) Therefore, Percentage of energy passed through the obstacle is given by E t E i × 100 = C ( 16 k 2 / 4 ) C ( 5 k / 2 ) 2 × 100 = 64 %

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