JEE Exam > JEE Questions > A stationary observer receives sonic oscillat...
Start Learning for Free
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:
- a)1 m/s
- b)2 m/s
- c)0.5 m/s
- d)1.5 m/s
Correct answer is option 'C'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Verified Answer
A stationary observer receives sonic oscillations from two tuning fork...
Most Upvoted Answer
A stationary observer receives sonic oscillations from two tuning fork...
To solve this problem, let's consider the situation from the perspective of the observer.
Given:
- Frequency of the tuning forks: 680 Hz
- Beat frequency heard by the observer: 2 Hz
- Velocity of sound in air: 340 m/s
Let's assume:
- The tuning fork approaching the observer has a frequency of f1.
- The tuning fork receding from the observer has a frequency of f2.
Using the concept of Doppler effect, we can relate the observed frequency with the actual frequency of the tuning forks.
The observed frequency when a source is moving towards the observer is given by:
f' = (v + v₀) / (v + vs) * f
where:
- f' is the observed frequency
- v is the velocity of sound in air
- v₀ is the velocity of the observer
- vs is the velocity of the source
- f is the actual frequency of the source
Similarly, the observed frequency when a source is moving away from the observer is given by:
f' = (v - v₀) / (v - vs) * f
In this case, the beat frequency is the difference between the observed frequencies of the two tuning forks:
Beat frequency = |f1' - f2'|
Since the observer is stationary, the velocity of the observer (v₀) is 0.
Now, let's calculate the observed frequencies for the two tuning forks:
For the approaching tuning fork:
f1' = (v + 0) / (v + vs) * f1
= (340) / (340 + vs) * 680
= (680 * 340) / (340 + vs)
For the receding tuning fork:
f2' = (v - 0) / (v - vs) * f2
= (340) / (340 - vs) * 680
= (680 * 340) / (340 - vs)
Now, substitute the values into the equation for beat frequency:
2 = |(680 * 340) / (340 + vs) - (680 * 340) / (340 - vs)|
Simplifying the equation:
2 = |(680 * 340 * (340 - vs) - 680 * 340 * (340 + vs)) / (340² - vs²)|
2 = |680 * 340 * (340 - vs + 340 + vs) / (340² - vs²)|
2 = |680 * 340 * (680) / (340² - vs²)|
2 = |680 * 680 / (340² - vs²)|
2 = |680 / (340 - vs)(340 + vs)|
Since the beat frequency cannot be negative, we can remove the absolute value:
2 = 680 / (340 - vs)(340 + vs)
Now, solve for vs by cross-multiplying:
2(340 - vs)(340 + vs) = 680
680 - 2vs² = 680
2vs² = 0
vs² = 0
vs = 0
Therefore, the speed of each tuning fork is 0 m/s.
However, it seems there might be an error in the question or answer options provided. It is not possible for both tuning forks to have a speed of 0 m/s and produce a beat frequency
Given:
- Frequency of the tuning forks: 680 Hz
- Beat frequency heard by the observer: 2 Hz
- Velocity of sound in air: 340 m/s
Let's assume:
- The tuning fork approaching the observer has a frequency of f1.
- The tuning fork receding from the observer has a frequency of f2.
Using the concept of Doppler effect, we can relate the observed frequency with the actual frequency of the tuning forks.
The observed frequency when a source is moving towards the observer is given by:
f' = (v + v₀) / (v + vs) * f
where:
- f' is the observed frequency
- v is the velocity of sound in air
- v₀ is the velocity of the observer
- vs is the velocity of the source
- f is the actual frequency of the source
Similarly, the observed frequency when a source is moving away from the observer is given by:
f' = (v - v₀) / (v - vs) * f
In this case, the beat frequency is the difference between the observed frequencies of the two tuning forks:
Beat frequency = |f1' - f2'|
Since the observer is stationary, the velocity of the observer (v₀) is 0.
Now, let's calculate the observed frequencies for the two tuning forks:
For the approaching tuning fork:
f1' = (v + 0) / (v + vs) * f1
= (340) / (340 + vs) * 680
= (680 * 340) / (340 + vs)
For the receding tuning fork:
f2' = (v - 0) / (v - vs) * f2
= (340) / (340 - vs) * 680
= (680 * 340) / (340 - vs)
Now, substitute the values into the equation for beat frequency:
2 = |(680 * 340) / (340 + vs) - (680 * 340) / (340 - vs)|
Simplifying the equation:
2 = |(680 * 340 * (340 - vs) - 680 * 340 * (340 + vs)) / (340² - vs²)|
2 = |680 * 340 * (340 - vs + 340 + vs) / (340² - vs²)|
2 = |680 * 340 * (680) / (340² - vs²)|
2 = |680 * 680 / (340² - vs²)|
2 = |680 / (340 - vs)(340 + vs)|
Since the beat frequency cannot be negative, we can remove the absolute value:
2 = 680 / (340 - vs)(340 + vs)
Now, solve for vs by cross-multiplying:
2(340 - vs)(340 + vs) = 680
680 - 2vs² = 680
2vs² = 0
vs² = 0
vs = 0
Therefore, the speed of each tuning fork is 0 m/s.
However, it seems there might be an error in the question or answer options provided. It is not possible for both tuning forks to have a speed of 0 m/s and produce a beat frequency
Attention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Top Courses for JEEView all
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer?
Question Description
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer?.
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A stationary observer receives sonic oscillations from two tuning forks, one of which approaches and the other recedes with same speed. As this takes place the observer hears the beat frequency of 2 Hz. Find the speed of each tuning fork, if their oscillation frequency is 680 Hz and the velocity of sound in air is 340 m/s:a)1 m/sb)2m/sc)0.5m/sd)1.5m/sCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup