Question

If the acoustic impedance of the metal is 4.99×107 Pa·s/m, what fraction of the initial intensity remains at the start of this second round trip?

Answer 1

Can you provide more context or, better, post the full question, please?

Answer 2

If it is a problem to do with a sound pulse bouncing backwards and forwards between the metal/air interfaces then you will also need the acoustic impedance, z, of air.

The you need to use the standard reflection formula as a fraction of reflected to incident intensities twice, one for each reflection.

\[\dfrac {I_r}{I_i} = \left( \dfrac{Z_1-Z_2}{Z_1+Z_2}\right)^2\]