Question

A girl is sitting near the open window of a train that is moving at a velocity of 10.00m/sto the east. The girl’s uncle stands near the tracks and watches the train move away. The locomotive whistle emits sound at frequency 500.0Hz. The air is still. (a) What frequency does the uncle hear? (b) What frequency does the girl hear? A wind begins to blow from the east at 10.00m/s. (c) What frequency does the uncle now hear? (d) What frequency does the girl now hear?

Short answer

1. The frequency heard bytheuncle when air is still is 485.8Hz.
2. The frequency heard bythegirl when air is still is 500Hz.
3. The frequency heard by the uncle when wind begins to blow from the east at 10.00m/s is 486Hz.
4. The frequency heard by the girl when wind begins to blow from the east at 10.00m/s is 500Hz.

Step-by-step solution

The given data

1. Velocity of the train and girl is V_g = V_t = 10.00m/s
2. Frequency of whistle is f = 500.0Hz

Understanding the concept of Doppler’s Effect

We can use the concept of the Doppler Effect. We use the equation of the Doppler Effect to find the frequency heard by the uncle and girl when the air is still. When air is moving, we find the relative speeds of the uncle and the girl with air and using those speeds, we can find the frequencies heard by them.

Formula:

The frequency received by the observer according to Doppler’s Effect,

$\mathit{f}\mathbf{\text{'}}\mathbf{=}\left(\frac{V\pm V_l}{V\pm V_s}\right)\mathit{f}$ …(i)

a) Calculation of the frequency heard by the uncle when air is still

Here the source is moving away from the uncle, so we can write the frequency using equation as:

$\begin{array}{rcl}f\text{'}& =& 500Hz\times \left(\frac{343m/s+0}{343m/s+10m/s}\right)\\ & =& 485.6Hz\end{array}$

Hence, the frequency heard by the uncle when air is still is 485.6Hz.

b) Calculation of the frequency heard by the girl when air is still

Here the girl is moving withthetrain. Hence, the relative speed ofthegirl with respect tothetrain is zero, so she will hear the same frequency. The girl will hear the frequency:

$\begin{array}{rcl}f\text{'}& =& f\\ & =& 500.0Hz\end{array}$

Hence, the frequency heard by the girl is 500.0Hz

c) Calculation of the frequency heard by the uncle when wind begins to blow from the east at 10.00m/s

Here, the speed of the train relative to air is V_t = 20m/s and the relative speed of the uncle with air is V_u = 10m/s so we can write the frequency using equation (i) as:

$\begin{array}{rcl}f\text{'}& =& 500Hz\times \left(\frac{343m/s+10m/s}{343m/s+20m/s}\right)\\ & =& 486Hz\end{array}$

Hence, the frequency heard by the uncle is 486Hz

d) Calculation of the frequency heard by the girl when wind begins to blow from the east at 10.00m/s

Here the girl is moving with the train, so the relative speed of girl with respect to air is V_g = 20m/s, and the speed of the train with air is also V_t = 20m/s Hence there is no relative motion between the girl andthe train, so the girl will hear the frequency:

$\begin{array}{rcl}f\text{'}& =& f\\ & =& 500.0Hz\end{array}$

Hence, the frequency heard by the girl is 500.0Hz.