[SOLVED] Comparison of sound waves in air and underwater 1. The problem statement, all variables and given/known data In air, the human ear is most sensitive to sound waves of 3300Hz. The auditory canal is part of the human ear. It is a tube of length 0.025m, i.e. 1/4 the wavelength of sound with frequency 3300Hz. When under water, the speed of sound is 1500m/s. The human ear becomes most sensitive to sounds with a frequency greater than that of the 3300Hz in air. Explain why this is. 2. Relevant equations and data Wavelength of sound in air = 4 x 0.025m = 0.1m Speed of sound = wavelength x frequency 3. The attempt at a solution The auditory canal is a closed tube, the eardrum being the closed end. Thus, the ear is most sensitive to sounds that strike the eardrum at a node: Under water, the ear will still be most sensitive to sounds of wavelength 4 times the length of the auditory canal, i.e. sound of wavelength 0.1m. However, speed of sound under water is greater so the frequency of the sound the ear is most sensitive to must be greater than 3300 Hz. Frequency x 0.1m = 1500m/s Frequency = 15000Hz The ear is thus most sensitive to frequencies of 15000Hz. ------------------------------------ This solution seems to go with the data provided. However, if the sound waves strike the eardrum at a node, shouldn't destructive interference occur? How is the person able to hear anything at all? Any light into this matter would be really appreciated.