As shown in the figure, a person is standing somewhere between two...

Given:
Frequency of tuning forks, $f = 512 \text{ Hz}$
Speed of tuning forks, $v_s = 5.5 \text{ m s}^{-1}$
Speed of sound, $v = 330 \text{ m s}^{-1}$

The tuning fork on the left moves towards the stationary listener. The apparent frequency heard by the listener is:
$f_1 = f \left( \dfrac{v}{v - v_s} \right)$

The tuning fork on the right moves away from the stationary listener. The apparent frequency heard by the listener is:
$f_2 = f \left( \dfrac{v}{v + v_s} \right)$

The number of beats heard by the listener is the difference between these two apparent frequencies:
$f_B = f_1 - f_2 = f \left( \dfrac{v}{v - v_s} - \dfrac{v}{v + v_s} \right)$

$f_B = fv \left( \dfrac{v + v_s - (v - v_s)}{v^2 - v_s^2} \right) = \dfrac{2 f v v_s}{v^2 - v_s^2}$

Since $v_s \ll v$, we can approximate $v^2 - v_s^2 \approx v^2$:
$f_B \approx \dfrac{2 f v v_s}{v^2} = \dfrac{2 f v_s}{v}$

Substituting the given values:
$f_B = \dfrac{2 \times 512 \times 5.5}{330}$

$f_B = \dfrac{512 \times 11}{330} = \dfrac{512}{30} \approx 17.067 \text{ Hz}$

Rounding to one decimal place, we get $17.1 \text{ Hz}$.