A train whistle is heard at 310Hz as the train approaches town. The train cuts its speed in half as it nears the station, and the sound of the whistle is then 300 Hz.

To explain why the frequency of the train whistle changes, we need to understand the concept of Doppler effect. The Doppler effect refers to the change in frequency of a sound wave due to the relative motion between the source of the sound and the observer.

In this case, the train whistle is initially heard at a frequency of 310 Hz as the train approaches the town. However, as the train gets closer to the observer (station), the distance between the observer and the train decreases. As a result, the waves of sound emitted by the whistle get compressed, leading to a higher frequency.

When the train cuts its speed in half as it nears the station, its relative velocity with respect to the observer decreases. This reduction in velocity causes the compression of sound waves to decrease, resulting in a lower frequency.

Now, let's explain how to calculate the frequency heard by the observer when the train cuts its speed in half. We can use the formula for Doppler effect, which is given as:

f' = f * (v + vo) / (v + vs)

Where:
f' is the observed frequency,
f is the actual frequency of the source (train whistle),
v is the speed of sound in air (assumed to be constant),
vo is the velocity of the observer (stationary),
and vs is the velocity of the source (train).

Since the velocity of the observer is zero (stationary), vo = 0. And we are given the initial frequency (f) as 310 Hz.

When the train cuts its speed in half, we can assume that the velocity of the source (vs) is reduced to half of its initial value. Let's say the initial velocity of the train was v0. The new velocity of the train near the station would be v0/2.

Plugging these values into the formula, we can calculate the new frequency (f') heard by the observer:

f' = 310 Hz * (v + 0) / (v + v0/2)
= 310 Hz * (v + v0/2) / (v + v0/2)

Simplifying the equation, we find that:

f' = 310 Hz * (v + v0/2) / (v + v0/2)
= 310 Hz

Therefore, the new frequency heard by the observer when the train cuts its speed in half is 300 Hz.