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Two pipes, equal in length, are each open at one end. Each has a fundamental frequency of 475 Hz at 297 K. In one pipe the air temperature is increased to 304 K. If the two pipes are sounded together, what beat frequency results?

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    The speed of sound is higher by a factor sqrt(475/297) = 1.265 at the higher temperature.

    Frequency = (wave speed)/(wavelength)
    Fundamental wavelength will not change with T because it depends upon pipe length.

    If the cooler-air pipe has a fundamental frequency of 475 Hz, it will be 475*1.265 = 600.1 Hz at the higher temperature.

    The beat frequency will be 600 - 475 = 125 Hz.

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