//ffλτμτ′′′′=¡=where we are making an assumption that the mass-per-unit-length of the string does not change significantly. Thus, with ′=1.2, we have/ 4401.2 ,f′=which gives482 Hzf′ =.(b) In this case, neither tension nor mass-per-unit-length change, so the wave speed vis unchanged. Hence, using Eq. 17–38 with
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This note was uploaded on 03/20/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.