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

The distance traveled by the tuning fork is 9.37 m

Step-by-step 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 [(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[(343)/(343 + v_s) ]\\\\(657)/(683) = (343)/(343 + v_s)\\\\0.962 = (343)/(343 + v_s)\\\\0.962(343 + v_s) = 343\\\\343 + v_s = (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 = (v_s^2)/(2g) \\\\h = (13.55^2 )/(2* 9.8) \\\\h = 9.37 \ m`

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