The pressure wave,P = 0.01 sin(1000t - 3x) , | vedclass

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(D) The given wave equation is $P = 0.01 \sin(1000t - 3x)$.Comparing this with the standard wave equation $P = A \sin(\omega t - kx)$,we get angular f

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$4$

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(D) The given wave equation is $P = 0.01 \sin(1000t - 3x)$.Comparing this with the standard wave equation $P = A \sin(\omega t - kx)$,we get angular frequency $\omega = 1000 \, rad/s$ and wave number $k = 3 \, m^{-1}$.The speed of sound $v_0$ at $0 \, ^oC$ $(T_0 = 273 \, K)$ is given by $v_0 = \frac{\omega}{k} = \frac{1000}{3} \approx 333.33 \, m s^{-1}$.We know that the speed of sound $v \propto \sqrt{T}$,where $T$ is the absolute temperature in Kelvin.Therefore,$\frac{v}{v_0} = \sqrt{\frac{T}{T_0}}$.Given $v = 336 \, m s^{-1}$ at temperature $T$,we have $\frac{336}{1000/3} = \sqrt{\frac{T}{273}}$.$\frac{1008}{1000} = \sqrt{\frac{T}{273}} \implies 1.008 = \sqrt{\frac{T}{273}}$.Squaring both sides,$1.016 = \frac{T}{273}$.$T = 273 	imes 1.016 = 277.368 \, K$.Converting to Celsius: $T( ^oC) = 277.368 - 273 = 4.368 \, ^oC$.The approximate value is $4 \, ^oC$.

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