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A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)
- a)1024
- b)512
- c)256
- d)128
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A tube, closed at one end containing air, produces, when excited, the...
For a closed tube, 
For an ipen tube, 
Hence v’ = 2v = 2 x 512 = 1024 Hz
Thus the correct choice is (a)
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Community Answer
A tube, closed at one end containing air, produces, when excited, the...
Explanation:
When a tube is closed at one end, it creates a node at the closed end and an antinode at the open end. This means that the tube can vibrate in a way that supports standing waves with a wavelength that is twice the length of the tube.
The fundamental frequency is the lowest frequency that can be produced in a system. In a tube closed at one end, the fundamental frequency is the frequency of the first harmonic, which corresponds to a single half-wavelength fitting inside the tube.
Given that the fundamental frequency is 512 Hz, we can determine the wavelength of the sound wave produced by the tube using the formula:
λ = 2L
Where λ is the wavelength and L is the length of the tube.
Since the tube is closed at one end, the length of the tube (L) is equal to half the wavelength (λ/2). Substituting this into the formula, we get:
λ = 2(L/2) = L
Therefore, the wavelength of the sound wave produced by the tube is equal to the length of the tube.
Now, let's consider a tube that is open at both ends. In this case, there will be a node at each end and an antinode in the middle. This means that the tube can vibrate in a way that supports standing waves with a wavelength that is equal to the length of the tube.
The fundamental frequency in a tube open at both ends corresponds to the second harmonic, where the wavelength of the sound wave is equal to the length of the tube (λ = L/2).
Since we know that the wavelength of the sound wave produced by the tube in the first case (closed at one end) is equal to the length of the tube, we can conclude that the fundamental frequency in the second case (open at both ends) will be twice the frequency of the first case.
Therefore, the fundamental frequency that can be excited in the tube open at both ends will be 2 * 512 Hz = 1024 Hz.
Conclusion:
The correct answer is option A) 1024.
When a tube is closed at one end, it creates a node at the closed end and an antinode at the open end. This means that the tube can vibrate in a way that supports standing waves with a wavelength that is twice the length of the tube.
The fundamental frequency is the lowest frequency that can be produced in a system. In a tube closed at one end, the fundamental frequency is the frequency of the first harmonic, which corresponds to a single half-wavelength fitting inside the tube.
Given that the fundamental frequency is 512 Hz, we can determine the wavelength of the sound wave produced by the tube using the formula:
λ = 2L
Where λ is the wavelength and L is the length of the tube.
Since the tube is closed at one end, the length of the tube (L) is equal to half the wavelength (λ/2). Substituting this into the formula, we get:
λ = 2(L/2) = L
Therefore, the wavelength of the sound wave produced by the tube is equal to the length of the tube.
Now, let's consider a tube that is open at both ends. In this case, there will be a node at each end and an antinode in the middle. This means that the tube can vibrate in a way that supports standing waves with a wavelength that is equal to the length of the tube.
The fundamental frequency in a tube open at both ends corresponds to the second harmonic, where the wavelength of the sound wave is equal to the length of the tube (λ = L/2).
Since we know that the wavelength of the sound wave produced by the tube in the first case (closed at one end) is equal to the length of the tube, we can conclude that the fundamental frequency in the second case (open at both ends) will be twice the frequency of the first case.
Therefore, the fundamental frequency that can be excited in the tube open at both ends will be 2 * 512 Hz = 1024 Hz.
Conclusion:
The correct answer is option A) 1024.
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A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)a)1024b)512c)256d)128Correct answer is option 'A'. Can you explain this answer?
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A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)a)1024b)512c)256d)128Correct answer is option 'A'. 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 A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)a)1024b)512c)256d)128Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)a)1024b)512c)256d)128Correct answer is option 'A'. Can you explain this answer?.
A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)a)1024b)512c)256d)128Correct answer is option 'A'. 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 A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)a)1024b)512c)256d)128Correct answer is option 'A'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A tube, closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, then the fundamental frequency that can be excited will be (in Hz)a)1024b)512c)256d)128Correct answer is option 'A'. Can you explain this answer?.
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