InterviewSolution
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InterviewSolution
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The equation of a sound wave in air is given by `rho = (0.01 N//m^(2)) [sin (1000 s^(-1)) t - (3.0 m^(-1))x]` (a) Find the frequency, wavelength and the speed of sound wave in air. (b) If the equilibrium pressure of air is `1.0 xx 10^(5) N//m^(2)`, what are the maximum and minimum pressure at a point as the wave passes through that point ? |
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Answer» `p = 0.01 sin (1000 t - 3 x)` (i) `p = p_(0) sin (omega t - k x)` (ii) Comparing (i) and (ii) `p_(0) = 0.01 N//m^(2), omega = 1000 rad//s, k = 3 m^(-1)` (a) `f = (omega)/(2 pi) = (1000)/(2 pi) = (500)/(pi) Hz` `k = (pi)/(lambda) rArr lambda = (2 pi)/(k) = (2 pi)/(3)m` `v = f lambda = (500)/(pi) xx (2 pi)/(3) = (1000)/(3) m//s` (b) `p_(max) = P_(0) + p_(0) = (1.01 xx 10^(5) + 0.01) N//m^(2)` `p_(min) = P_(0) - p_(0) = (1.01 xx 10^(5) - 0.01) N//m^(2)` `P_(0)`: atmospheric pressure |
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