Answer:
Considering the two cases of when the train is moving towards and away from the stationary observer. The observed frequency of sound waves from the train can decrease when
1) The speed of the train is slowly reducing and its direction is towards the stationary observer.
2) The speed of the train is slowly increasing and its direction is away from the observer for the sound's observed frequency to keep decreasing.
Step-by-step explanation:
This phenomenon is due to Doppler's Effect.
Doppler's Effect explains how relative frequency of a sound source varies with the velocity of the source or the observer.
Generally, the mathematical expression for Doppler's Effect is given below
f' = f [(v + v₀)/(v - vₛ)]
where
f' = observed frequency
f = actual frequency
v = velocity of sound waves
v₀ = velocity of observer
vₛ = velocity of sound source
When the train is moving towards the stationary observer,
f' = observed frequency of the sound wave = f₁
f = actual frequency of the sound wave = f
v = velocity of sound waves = v
v₀ = velocity of observer = 0 m/s
vₛ = velocity of sound source = vₛ
f' = f [(v + v₀)/(v - vₛ)]
f₁ = f [(v + 0)/(v - vₛ)]
f₁ = fv/(v - vₛ) (eqn 1)
Observed frequency is obviously higher than the frequency of the train.
But if the train keeps reducing its speed (vₛ) towards from the stationary observer, the observed frequency will decrease.
When the train is moving away from the stationary observer,
f' = observed frequency of the sound wave = f₂
f = actual frequency of the sound wave = f
v = velocity of sound waves = v
v₀ = velocity of observer = 0 m/s
vₛ = velocity of sound source = - vₛ (train is moving away from the observer, hence, the negative sign)
f' = f [(v + v₀)/(v - vₛ)]
f₂ = f [(v + 0)/(v - (-vₛ)]
f₂ = fv/(v + vₛ) (eqn 2)
From the expression, it is clear that the observed frequency is smaller than the frequency of the sound waves and it keeps decreasing as the speed of the train (vₛ) increases as the train moves away from the stationary observer.
Hope this Helps!!!