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?

Answer 1

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.