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The equation of a sound wave in air is given by `trianglep=(0.02)sin[(3000)t-(9.0)x]`, where all variables are is SI units. (a) find the frequency, wavelength and the speed of sound wave in air. (b) If the equilibrium pressure of air is `1.01xx10^5(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» a. comparing with the standard form of a travelling wave, `trianglep=trianglep_(max)sin[omegat-(x)/(v)]` we see that `omega=3000s^-1.` The frequency is `f=(omega)/(2pi)=(3000)/(2pi)Hz` Also from the same comparison, `(omega)/(v)=9.0m^-1` or `v=(omega)/(9.0m^-1)=(3000s^-1)/(9.0m^-1)=(1000)/(3)(m)/(s)` The wavelength is `lamda=(v)/(f)=((1000)/(3(m)/(s)))((3000)/(2piHz))=(2pi)/(9)m` b. The pressure amplitude is `trianglep_(max)=0.02(N)/(m^2)`. Hence the maximum and minimum pressures at a point in the wave motion will be `1.01xx10^5+-0.02(N)/(m^2)` |
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