Question 15.2: The amplitude of a vibrating member decreases so that the am......

\Lambda =\ln (e_{n}/e_{n+1} ) . Since (e_{m}/e_{m+1} )= (e_{n}/e_{n+1} )= (e_{n+1}/e_{n+2} ), and so forth, and (e_{n}/e_{n+m} )= (e_{n}/e_{n+1} )(e_{n+1}/e_{n+2} )(e_{n+2}/e_{n+3})…(e_{n+m-1}/e_{n+m})= m(e_{n}/e_{n+m}). Therefore (e_{n}/e_{n+1} )= (e_{n}/e_{n+m} )^{1/m}. \ \Lambda =\ln (e_{n}/e_{n+1} ) =\ln (e_{n}/e_{m} ) ^{1/m}  =(1/m)\ln (e_{n}/e_{n+m}). Substituting m = 100 and (e_{n}/e_{n+m} )=1/0.13, \ \Lambda =0.0204.