Question

NAME A tuning fork vibrating at 512 Hz falls from rest and How far below the point of release is the tuning fork when waves of frequency 485 Hz reach the release point? Assume the speed of sound is 343 m/s. as a result of gravity

Answer 1

Here source of sound is moving away from the observer. Let the fuming fork is at a distance d from the release point when it's frequency is heard to be 485 Hz. Then it's velocity at that point is V = squareroot 2g d where g is acceleration due to gravity. The apparent frequency is given by 485 = squareroot 2 g d/V_s + squareroot 2g d times 512 [r^' = V/V_s + V r_0, where V_s is speed of sound] Thus, we have V_s + squareroot 2g d/squareroot 2gd = 512/485 V_s/squareroot 2gd = 512/485 - 1 = 27/485 squareroot 2gd = 27/485 times V_s alpha = (27/485 times V_s)^2 times 2/2g = (27/485 times 343)^2 times 1/2 times 9.8 = 18.60 m (below)