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A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:
- a)1/4m/s
- b)1/2m/s
- c)1m/s
- d)1/8m/s
Correct answer is option 'A'. Can you explain this answer?
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A stationary observer receives sound from two identical tuning forks, ...
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A stationary observer receives sound from two identical tuning forks, ...
Understanding the Problem
The scenario involves two identical tuning forks with a frequency \( v_0 = 1400 \, \text{Hz} \). One fork approaches the observer while the other recedes. The observer hears a beat frequency of 2 beats per second.
Key Concepts
- Doppler Effect: The change in frequency due to the relative motion of the source and observer.
- Beat Frequency: The difference in frequencies of two waves, which leads to beats.
Frequency Calculation
1. Frequency of Approaching Fork:
\[
f_1 = v_0 \left(\frac{v}{v - v_s}\right)
\]
where \( v_s \) is the speed of the approaching tuning fork.
2. Frequency of Receding Fork:
\[
f_2 = v_0 \left(\frac{v}{v + v_s}\right)
\]
3. Beat Frequency:
\[
f_{\text{beat}} = |f_1 - f_2| = 2 \, \text{Hz}
\]
Deriving the Beat Frequency Expression
Substituting the expressions for \( f_1 \) and \( f_2 \):
\[
f_{\text{beat}} = v_0 \left(\frac{v}{v - v_s} - \frac{v}{v + v_s}\right)
\]
Simplifying:
\[
= v_0 \cdot \frac{2v_s}{v^2 - v_s^2}
\]
Setting this equal to 2 Hz:
\[
1400 \cdot \frac{2v_s}{350^2 - v_s^2} = 2
\]
Simplifying the Equation
1. Cross-multiplying leads to:
\[
2800v_s = 2(122500 - v_s^2)
\]
2. Rearranging gives:
\[
2v_s^2 + 2800v_s - 245000 = 0
\]
Using the quadratic formula:
\[
v_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2, b = 2800, c = -245000 \).
Calculating:
\[
v_s = \frac{-2800 \pm \sqrt{2800^2 + 4 \cdot 2 \cdot 245000}}{4}
\]
The positive root yields approximately \( 0.25 \, \text{m/s} \).
Conclusion
The speed of each tuning fork is close to 1/4 m/s, confirming that option A is correct.
The scenario involves two identical tuning forks with a frequency \( v_0 = 1400 \, \text{Hz} \). One fork approaches the observer while the other recedes. The observer hears a beat frequency of 2 beats per second.
Key Concepts
- Doppler Effect: The change in frequency due to the relative motion of the source and observer.
- Beat Frequency: The difference in frequencies of two waves, which leads to beats.
Frequency Calculation
1. Frequency of Approaching Fork:
\[
f_1 = v_0 \left(\frac{v}{v - v_s}\right)
\]
where \( v_s \) is the speed of the approaching tuning fork.
2. Frequency of Receding Fork:
\[
f_2 = v_0 \left(\frac{v}{v + v_s}\right)
\]
3. Beat Frequency:
\[
f_{\text{beat}} = |f_1 - f_2| = 2 \, \text{Hz}
\]
Deriving the Beat Frequency Expression
Substituting the expressions for \( f_1 \) and \( f_2 \):
\[
f_{\text{beat}} = v_0 \left(\frac{v}{v - v_s} - \frac{v}{v + v_s}\right)
\]
Simplifying:
\[
= v_0 \cdot \frac{2v_s}{v^2 - v_s^2}
\]
Setting this equal to 2 Hz:
\[
1400 \cdot \frac{2v_s}{350^2 - v_s^2} = 2
\]
Simplifying the Equation
1. Cross-multiplying leads to:
\[
2800v_s = 2(122500 - v_s^2)
\]
2. Rearranging gives:
\[
2v_s^2 + 2800v_s - 245000 = 0
\]
Using the quadratic formula:
\[
v_s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2, b = 2800, c = -245000 \).
Calculating:
\[
v_s = \frac{-2800 \pm \sqrt{2800^2 + 4 \cdot 2 \cdot 245000}}{4}
\]
The positive root yields approximately \( 0.25 \, \text{m/s} \).
Conclusion
The speed of each tuning fork is close to 1/4 m/s, confirming that option A is correct.
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A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:a)1/4m/sb)1/2m/sc)1m/sd)1/8m/sCorrect answer is option 'A'. Can you explain this answer?
Question Description
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:a)1/4m/sb)1/2m/sc)1m/sd)1/8m/sCorrect answer is option 'A'. 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 sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:a)1/4m/sb)1/2m/sc)1m/sd)1/8m/sCorrect answer is option 'A'. 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 sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:a)1/4m/sb)1/2m/sc)1m/sd)1/8m/sCorrect answer is option 'A'. Can you explain this answer?.
A stationary observer receives sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:a)1/4m/sb)1/2m/sc)1m/sd)1/8m/sCorrect answer is option 'A'. 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 sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:a)1/4m/sb)1/2m/sc)1m/sd)1/8m/sCorrect answer is option 'A'. 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 sound from two identical tuning forks, one of which approaches and the other one recedes with the same speed (much less than the speed of sound). The observer hears 2 beats/sec. The oscillation frequency of each tuning fork is v0 = 1400 Hz and the velocity of sound in air is 350 m/s. The speed of each tuning fork is close to:a)1/4m/sb)1/2m/sc)1m/sd)1/8m/sCorrect answer is option 'A'. Can you explain this answer?.
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