Question

A TRAIN SOUNDS ITS WHISTLE AS IT APPROACHES AND LEAVES A RAILROAD CROSSING. AN OBSERVER AT THE CROSSING MEASURES A FREQUENCY OF 219 HZ AS THE TRAIN APPROACHES AND A FREQUENCY OF 184 HZ AS THE TRAIN LEAVES. THE SPEED OF SOUND IS 340 M/S. FIND THE SPEED OF THE TRAIN AND THE FREQUENCY OF ITS WHISTLE. .

Please provide work shown and show what each variable represents thank you so much.

Answer 1

We use,

- Observed Frequency

- Actual Frequency

- Velocity of Sound = 340 m/s

$v_t$ - Velocity of Train

Doppler Equation for Stationary Observer & Approaching Source,  $f'_1 = f \left [ \frac{v}{v - v_t} \right ]$

Substituting, $219 = f \left [ \frac{340}{340 - v_t} \right ] \rightarrow Eqn (1)$

Doppler Equation for Stationary Observer & Leaving Source,  $f'_2 = f \left [ \frac{v}{v + v_t} \right ]$

Substituting, $184 = f \left [ \frac{340}{340 + v_t} \right ] \rightarrow Eqn (2)$

$\frac{Eqn (1)}{Eqn (2) } = \frac{219}{184} = \frac{340 + v_t }{ 340 - v_t}$

Cross multiplying, $\Rightarrow (219*340) - 219v_t = (340*184) + 184v_t$

$\Rightarrow 403v_t = 11900 \Rightarrow v = \frac{11900}{403} = 29.53 m/s \rightarrow Answer$

Velocity of Train is 29.53 m/s

So Eqn (1) becomes  $219 = f \left [ \frac{340}{340 - 29.52} \right ]$

Re arranging, $\Rightarrow f = \frac{219*310.48}{340} =199.98 Hz\rightarrow Answer$

Actual Frequency of Train is 199.98 Hz