An observer is standing on the platform of a railway station.A train goes through the station without stopping.If the frequency of the train whistle decreases by a factor of 1.2 as it approaches and then passes him,calculate the speed of the train (assume the speed of sound in air = 343 m/s).
A)112 km/hr
B)31.2 km/hr
C)56.0 km/hr
D)62.0 km/hr
E)122 km/hr

Question

Quizplus · Accepted Answer

The observed frequency change due to the Doppler effect can be calculated using the formula for the observed frequency when the source is moving towards and then away from the observer. The formula for the Doppler effect when the source is moving towards the observer is $f' = f \left(\frac{v + v_o}{v - v_s}\right)$, and when moving away it is $f' = f \left(\frac{v - v_o}{v + v_s}\right)$, where $f'$ is the observed frequency, $f$ is the actual frequency emitted by the source, $v$ is the speed of sound in the medium, $v_o$ is the speed of the observer (which is 0 in this case since the observer is stationary), and $v_s$ is the speed of the source (train).Given that the frequency decreases by a factor of 1.2, we can set up the ratio of the frequencies when the train is approaching and when it is receding as $1.2 = \frac{v + v_o}{v - v_s} \cdot \frac{v - v_o}{v + v_s}$. Since $v_o = 0$, this simplifies to $1.2 = \frac{v}{v - v_s} \cdot \frac{v}{v + v_s}$.Solving for $v_s$, we find that $v_s = v \left(\frac{1 - \sqrt{1.2}}{1 + \sqrt{1.2}}\right)$. Plugging in $v = 343$ m/s, we find $v_s \approx 31.1$ m/s, which is approximately 112 km/hr when converted from m/s to km/hr.

An Observer Is Standing on the Platform of a Railway