 # the quetion is mentioned in the image.i n the stationary wave pattern formed as a resdult of reflection from an obstacle the ratio of amplitude at antinode and node is 1.5.find the energy percent transmitted through the obstacle.

4 years ago
Energy in a wave = 0.5 *density *(amplitude*angular frequency )^2 . The wave reflected by the obstacle has an ampitude of (1/5) times the initial amplitude .Therefore energy transmitted through the wall is 0.5*density*(angular frequency*initial amplitude*4/5)^2
4 years ago
Dear student

The incident wave can be reflected or transmitted. (I think we must assume that none is absorbed or scattered.) The reflected wave interferes with the incident wave, producing a standing wave. But the reflected wave is weaker than the incident wave - ie the amplitude is smaller. So there is not complete constructive and destructive interference. The nodes are not zero amplitude.

If the incident wave has amplitude A and reflected wave has amplitudd of rA.

Then they generate stabding wave of amplitude (1+r)A at antinodes and (1-r)A at nodes.

We are given that

(1+r)A/ (1-r)A = 1.5 = 3/2

r = 1/5

The energy in the wave is proportional to amplitude squared, so the fraction of incident energy in the reflected wave is (1/5)² = 1/25 = 4%

The fraction of energy in the transmitted wave is (100-4) = 96%

Regards