InterviewSolution
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InterviewSolution
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The pressure variation in a sound wave is given by `DeltaP=8cos(4.00x-3000t+(pi)/(4))` Find its displacement amplitude. The density of the medium is `10^(3)" kg "m^(-3)`. |
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Answer» `DeltaP_(0)=BAK` Also, `B=rhov^(2)` `thereforeDeltaP_(0)=rhoV^(2)Ak` But `V^(2)=(omega^(2))/(k^(2))` `thereforeDeltaP_(0)=(rho omega^(2))/(k^(2))Ak=(rho omega^(2)A)/(k)` or `A=((DeltaP_(0))k)/(rho omega^(2))=(8xx4.00)/(10^(3)xx(3000)^(2))=3.55xx10^(-9)m` |
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