BIO Audible Sound. Provided the amplitude is sufficiently great, the human ear can respond to longitudinal waves over a range of frequencies from about 20.0 Hz to about 20.0 kHz. (a) If you were to mark the beginning of each complete wave pat-tern with a red dot for the long-wavelength sound and a blue dot for the short-wavelength sound, how far apart would the red dots be, and how far apart would the blue dots be? (b) In reality would adjacent dots in each set be far enough apart for you to easily measure their separation with a meter stick? (c) Suppose you repeated part (a) in water, where sound travels at 1480 m/s. How far apart would the dots be in each set? Could you readily measure their separation with a meter stick?

Solution 2E Problem (a) Step 1: Low frequency f L = 20.0 Hz High frequency f H = 20.0 KHz Step 2: Distance of separation is found using wavelengths. v We know that = f For shorter wavelength ( ), frequency is higher S For longer wavelength ( ),Lfrequency is lower Step 3: Red dot is marked for longer wavelength ( ) of low frequency (f ) L L f = 20.0 Hz L v L fL Where v is the velocity sound in air, v = 344 m/s = 344 L 20 L 17.2 m Red dots would be at 17.2 m Step 4: Blue dot is marked for shorter wavelength ( ) of high frequency(f ) S H f H = 20.0 KHz v S f H = 344 S 20*1000 S 0.0172 m or 1.72 cm Blue dots would be at 1.72 cm Problem (b) The separation of red and blue dots can be measured using meter stick. Problem (c) Step 1: To calculate the distance of red and blue dots if air is replaced by water Speed of sound in air v = 1480 m/s Step 2: Distance of red dot v L fL = 1480 L 20 L 74 m