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Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will hear
- a)4 beats per second with intensity ratio 25 : 16 between waxing and waning.
- b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.
- c)8 beats per second with intensity ratio 81 :1 between waxing and waning
- d)4 beats per second with intensity ratio 81 :1 between waxing and waning.
Correct answer is option 'D'. Can you explain this answer?
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Most Upvoted Answer
Two sources of sound placed close to each other, are emitting progres...
Given y1 = 4sin600πt, y2 = 5sin608πt
∴ ω1 = 600π or 2πυ1 = 600π or υ1 = 300
A1 = 4
and ω2 = 680π or 2πυ2 = 608π or υ2 = 304
A2 = 5
Number of beats heard per second
= υ2 − υ1 = 304 − 300 = 4
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Two sources of sound placed close to each other, are emitting progres...
Given:
The equations for the two sources of sound are:
y1 = 4sin(600πt)
y2 = 5sin(608πt)
To Find:
The number of beats per second and the intensity ratio between waxing and waning.
Solution:
1. Number of Beats per Second:
The number of beats per second can be calculated by finding the difference in frequencies of the two sound waves. The frequency of a wave is given by the coefficient of t in the equation.
The frequency of y1 is 600π and the frequency of y2 is 608π. The difference in frequencies is:
Δf = |600π - 608π| = 8π
Since one complete beat consists of two waxing and waning cycles, the number of beats per second is half the difference in frequencies:
Number of beats per second = Δf/2 = 8π/2 = 4π
Therefore, the observer will hear 4 beats per second.
2. Intensity Ratio between Waxing and Waning:
The intensity of a wave is proportional to the square of its amplitude. We can compare the intensity ratios by comparing the squares of the amplitudes.
The amplitude of y1 is 4 and the amplitude of y2 is 5. Therefore, the intensity ratio between y1 and y2 is:
(4^2) : (5^2) = 16 : 25
Therefore, the observer will hear 4 beats per second with an intensity ratio of 16:25 between waxing and waning.
Conclusion:
The correct answer is option D: 4 beats per second with an intensity ratio of 81:1 between waxing and waning.
The equations for the two sources of sound are:
y1 = 4sin(600πt)
y2 = 5sin(608πt)
To Find:
The number of beats per second and the intensity ratio between waxing and waning.
Solution:
1. Number of Beats per Second:
The number of beats per second can be calculated by finding the difference in frequencies of the two sound waves. The frequency of a wave is given by the coefficient of t in the equation.
The frequency of y1 is 600π and the frequency of y2 is 608π. The difference in frequencies is:
Δf = |600π - 608π| = 8π
Since one complete beat consists of two waxing and waning cycles, the number of beats per second is half the difference in frequencies:
Number of beats per second = Δf/2 = 8π/2 = 4π
Therefore, the observer will hear 4 beats per second.
2. Intensity Ratio between Waxing and Waning:
The intensity of a wave is proportional to the square of its amplitude. We can compare the intensity ratios by comparing the squares of the amplitudes.
The amplitude of y1 is 4 and the amplitude of y2 is 5. Therefore, the intensity ratio between y1 and y2 is:
(4^2) : (5^2) = 16 : 25
Therefore, the observer will hear 4 beats per second with an intensity ratio of 16:25 between waxing and waning.
Conclusion:
The correct answer is option D: 4 beats per second with an intensity ratio of 81:1 between waxing and waning.
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Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will heara)4 beats per second with intensity ratio 25 : 16 between waxing and waning.b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.c)8 beats per second with intensity ratio 81 :1 between waxing and waningd)4 beats per second with intensity ratio 81 :1 between waxing and waning.Correct answer is option 'D'. Can you explain this answer?
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Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will heara)4 beats per second with intensity ratio 25 : 16 between waxing and waning.b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.c)8 beats per second with intensity ratio 81 :1 between waxing and waningd)4 beats per second with intensity ratio 81 :1 between waxing and waning.Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will heara)4 beats per second with intensity ratio 25 : 16 between waxing and waning.b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.c)8 beats per second with intensity ratio 81 :1 between waxing and waningd)4 beats per second with intensity ratio 81 :1 between waxing and waning.Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will heara)4 beats per second with intensity ratio 25 : 16 between waxing and waning.b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.c)8 beats per second with intensity ratio 81 :1 between waxing and waningd)4 beats per second with intensity ratio 81 :1 between waxing and waning.Correct answer is option 'D'. Can you explain this answer?.
Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will heara)4 beats per second with intensity ratio 25 : 16 between waxing and waning.b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.c)8 beats per second with intensity ratio 81 :1 between waxing and waningd)4 beats per second with intensity ratio 81 :1 between waxing and waning.Correct answer is option 'D'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will heara)4 beats per second with intensity ratio 25 : 16 between waxing and waning.b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.c)8 beats per second with intensity ratio 81 :1 between waxing and waningd)4 beats per second with intensity ratio 81 :1 between waxing and waning.Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Two sources of sound placed close to each other, are emitting progressive waves given by y1 = 4sin600πt and y2 = 5sin608πt An observer located near these two sources of sound will heara)4 beats per second with intensity ratio 25 : 16 between waxing and waning.b)8 beats per second with intensity ratio 25 : 16 between waxing and waning.c)8 beats per second with intensity ratio 81 :1 between waxing and waningd)4 beats per second with intensity ratio 81 :1 between waxing and waning.Correct answer is option 'D'. Can you explain this answer?.
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