Question

A loudspeaker sounds a 610 Hz tone. What is the difference in the wavelength of the sound waves produced in air at 8

Answer 1

The speed ofsound with temperature is

        V = 331 + 0.606Tc

Speed ofsound at -6 is

       V = 331+ 0.606(8)

           = 335.848 m/s

Wavelength

         ? = V/f

            = [ 335.848 m/s] / [610 Hz]

             = 0.55057 m

Speed ofsound at 38 is

        V = 331+ 0.606(38)

           = 354.028 m/s

  Wavelength

          ? = V/f

             = [354.028 m/s] / [610 Hz]

            = 0.58037 m

the differencein wavelength is,

    change in wavelength = 0.58037 m - 0.55057 m

             = 0.0298 m

            = 2.98 cm

            = 3 cm