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A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).
Correct answer is '200'. Can you explain this answer?
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Verified Answer
A train sounds its whistle as it approaches and leaves a railroad cros...
Only the source is moving with respect to the medium. Let |vT| denote the speed of the train and v0 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)
...(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)Most Upvoted Answer
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.
- 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.
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Community Answer
A train sounds its whistle as it approaches and leaves a railroad cros...
Only the source is moving with respect to the medium. Let |vT| denote the speed of the train and v0 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)
...(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 crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer?
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A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer?.
A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer? for Physics 2024 is part of Physics preparation. The Question and answers have been prepared according to the Physics exam syllabus. Information about A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer? covers all topics & solutions for Physics 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer?.
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A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer?, a detailed solution for A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer? has been provided alongside types of A train sounds its whistle as it approaches and leaves a railroad crossing. An observer at the crossing measures a frequency of 219 Hz as the train approaches and a frequency of 184 Hz as the train leaves. The speed of sound is 340 m/s. Then calculate the frequency of its whistle (in Hz).Correct answer is '200'. Can you explain this answer? theory, EduRev gives you an
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