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A sound wave of frequency 10 kHz is travellilng in air with a speed of `340 ms^-1`. Find the minimum separation between two points where the phase difference is `60^@`. |
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Answer» The wavelength of the wave is `lamda=v/v=(340ms^-1)/(10xx10^3s^-1)=3.4cm` The wave number is `k=(2pi)/lamda=(2pi)/3.4cm^-1` The phase of the wave is `(kx-omegat)`. At any given instant the phase difference between two points at separatin d is kd. If this phase difference is `60^@` i.e. pi/3` radian, `pi/3=((2pi)/3.4cm^-1)d or d=3.4/6cm=0.57cm. |
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