A train sounds its whistle as it approaches and leaves a railroad cros...
Only the source is moving with respect to the medium. Let |v
T| denote the speed of the train and v
0 denote the emitted frequency, the observer measures a frequency
...(i)
as the train approaches, and a frequency
...(ii)
as the train recedes from him. Dividing Eq. (1) by Eq. (2), we find that
...(iii)
Solving this equation for the speed of the train we have
...(iv)
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A train sounds its whistle as it approaches and leaves a railroad cros...
Given information:
- Frequency of the train whistle as it approaches the crossing: 219 Hz
- Frequency of the train whistle as it leaves the crossing: 184 Hz
- Speed of sound: 340 m/s
Approach:
To solve this problem, we can use the Doppler effect formula for sound waves. The formula relates the observed frequency of a sound wave to the frequency of the source and the relative velocity between the source and the observer.
Doppler effect formula:
f_observed = (v_sound + v_observer) / (v_sound + v_source) * f_source
where,
f_observed = observed frequency
v_sound = speed of sound
v_observer = velocity of the observer
v_source = velocity of the source
f_source = frequency of the source
Analysis:
When the train approaches the crossing, the observer hears a higher frequency than the actual frequency of the whistle. This is because the sound waves are compressed due to the relative motion of the source (train) towards the observer. On the other hand, when the train leaves the crossing, the observer hears a lower frequency than the actual frequency of the whistle. This is because the sound waves are stretched due to the relative motion of the source (train) away from the observer.
Solution:
Let's assume the frequency of the train whistle is f_whistle.
When the train approaches the crossing:
f_observed_approaching = 219 Hz
v_observer_approaching = 0 m/s (since the observer is stationary)
v_source_approaching = v_train (velocity of the train)
Using the Doppler effect formula:
219 = (340 + 0) / (340 + v_train) * f_whistle
When the train leaves the crossing:
f_observed_leaving = 184 Hz
v_observer_leaving = 0 m/s
v_source_leaving = -v_train (negative velocity since the train is moving away)
Using the Doppler effect formula:
184 = (340 + 0) / (340 - v_train) * f_whistle
Solving the equations:
We have two equations with two unknowns (f_whistle and v_train). By solving these equations simultaneously, we can find the value of f_whistle, which is the frequency of the train whistle.
Solving the equations gives f_whistle = 200 Hz.
Conclusion:
Therefore, the frequency of the train whistle is 200 Hz.
A train sounds its whistle as it approaches and leaves a railroad cros...
Only the source is moving with respect to the medium. Let |v
T| denote the speed of the train and v
0 denote the emitted frequency, the observer measures a frequency
...(i)
as the train approaches, and a frequency
...(ii)
as the train recedes from him. Dividing Eq. (1) by Eq. (2), we find that
...(iii)
Solving this equation for the speed of the train we have
...(iv)