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The pressure variation in a sound wave in air is given by `Deltap = 12 sin (8.18X - 2700 t + pi//4) N//m^(2)` find the displacement amplitude. Density of air `= 1.29 kg//m^(3)`. |
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Answer» Correct Answer - A::D In the above problem, we have found that `A = (Delta p_(max))/(2 pi f rho nu)` `= ((Delta p_(max)k))/(rho omega^(2))` (as `nu = (omega)/(k)` and `2 pi f = omega`) Now subsituting the value, we have `A= ((12)(8.18))/((1.29)(2700)^(2)) = 1.04 xx 10^(-5)m` |
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