A violin player travelling on a slow train sounds a 440 Hz note,...

Let $f = 440 \text{ Hz}$ be the actual frequency of the note.

Let $v$ be the speed of sound and $v_s$ be the speed of the train.

For the observer on the ground, the source (train) is moving towards them. The apparent frequency heard by the person on the ground is:

f_1 = f \left(\dfrac{v}{v - v_s} \right)

The beat frequency detected by the ground observer is:

\Delta f_1 = f_1 - f = f \left(\dfrac{v}{v - v_s} -1 \right) = f \left(\dfrac{v_s}{v - v_s} \right) =$4.0 \text{ Hz}$

For the observer on the train, the source on the ground is stationary and the observer is moving towards it. The apparent frequency heard by the player on the train is:

$f_2 = f \left( \dfrac{v + v_s}{v} \right)$

The beat frequency detected by the train observer is:

\Delta f_2 = f_2 - f = f \left(\dfrac{v + v_s}{v} -1 \right) = f \left(\dfrac{v_s}{v} \right)

Comparing $\Delta f_1$ and $\Delta f_2$:

\Delta f_2 = f \left(\dfrac{v_s}{v - v_s} \right) \times \left( \dfrac{v - v_s}{v} \right) = \Delta f_1 \left( \dfrac{v - v_s}{v} \right)$<br><br>$\Delta f_2 = 4.0 \left( 1 - \dfrac{v_s}{v} \right)<br><br>Since the train is moving,v_s > 0, which implies \left( 1 -\dfrac{v_s}{v} \right)$< 1$.

Therefore, $\Delta f_2 < 4.0 \text{ Hz}$.

The beat frequency detected by the player on the train is a little less than $4.0 \text{ beats/s}.$