Question

A tuning fork vibrating at 508 Hz falls from rest and accelerates at 9.80 m/s^2. How far below the point of release is the tuning fork when waves of frequency of 490 Hz reach the release point? (Take the speed of sound in air to be 343 m/s.)

Answer 1

Answer:

Explanation:

given,

tuning fork vibration = 508 Hz

accelerates = 9.80 m/s²

speed of sound = 343 m/s

observed frequency = 490 Hz

$f_s = f(\dfrac{v}{v-(-v_s)})$

$f_s = f(\dfrac{v}{v+v_s})$

$v_s = v[\dfrac{f_s}{f_o}-1]$

      $= 343[\dfrac{508}{490}-1]$

      $v_s=12.6 m/s$

distance the tunning fork has fallen

$y_1=\dfrac{v^2}{2a_y}$

     $=\dfrac{12.6^2}{2\times 9.8}$

     =8.1 m

now, time required for the observed will be

$t = \dfrac{8.1}{343} = 0.023 s$

now, for the distance calculation

$y_2 = u\ t + \dfrac{1}{2}at^2$

  $= 12.6\times 0.023 +\dfrac{1}{2}\times 9.8 \times 0.023^2$

  =0.293 m

total distance

 = 8.1 + 0.293 = 8.392 m