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Navneet kumar
37 Points
10 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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