Question

You set a tuning fork into vibration at a frequency of 683 Hz and then drop it off the roof of the Physics building where the acceleration due to gravity is 9.80 m/s2. Determine how far the tuning fork has fallen when waves of frequency 657 Hz reach the release point

Answer 1

Answer:

The distance traveled by the tuning fork is 9.37 m

Explanation:

Given;

source frequency, $f_s$ = 683 Hz

observed frequency, $f_o$ = 657 Hz

The speed at which the tuning fork fell is calculated by applying Doppler effect formula;

$f_o = f_s [\frac{v}{v + v_s} ]$

where;

$v$ is speed of sound in air = 343 m/s

$v_s$ is the speed of the falling tuning fork

$657 = 683[\frac{343}{343 + v_s} ]\\\\\frac{657}{683} = \frac{343}{343 + v_s}\\\\0.962 = \frac{343}{343 + v_s}\\\\0.962(343 + v_s) = 343\\\\343 + v_s = \frac{343}{0.962} \\\\343 + v_s = 356.55\\\\v_s = 356.55 - 343\\\\v_s = 13.55 \ m/s$

The distance traveled by the tuning fork is calculated by applying kinematic equation as follows;

$v_s^2 = v_o^2 + 2gh$

where;

$v_o$ is the initial speed of the tuning fork = 0

g is acceleration due to gravity = 9.80 m/s²

$v_s^2 = 0 + 2gh\\\\h = \frac{v_s^2}{2g} \\\\h = \frac{13.55^2 }{2\times 9.8} \\\\h = 9.37 \ m$

Therefore, the distance traveled by the tuning fork is 9.37 m