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An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be?
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An open resonating tube has fundamental frequency of n when half of it...
Resonance in an Open Tube:
When a tube is open at both ends, it can resonate at certain frequencies, known as its harmonics. The fundamental frequency, also known as the first harmonic, is the lowest frequency at which the tube can resonate. This frequency is determined by the length of the tube.
Effect of Water on the Length of the Tube:
When half of the length of the tube is dipped into water, the effective length of the tube changes. The water acts as a barrier, effectively closing one end of the tube. This means that the tube is no longer open at both ends, but rather open at one end and closed at the other.
Change in the Fundamental Frequency:
The fundamental frequency of the open tube is given by the equation:
f = v / (2L)
Where f is the frequency, v is the speed of sound, and L is the length of the tube. When the tube is dipped into water, the effective length of the tube changes to half of its original length. Let's denote this new length as L'.
The new fundamental frequency can be calculated using the equation:
f' = v / (4L')
Since L' is half of the original length, we can substitute L/2 for L' in the equation:
f' = v / (4 * L/2)
= v / (4 * L/2)
= v / (2L)
Conclusion:
From the equation above, we can see that the fundamental frequency of the tube remains unchanged when half of its length is dipped into water. This is because the effective length of the tube is halved, but the relationship between the frequency and length remains the same.
Therefore, the fundamental frequency of the open resonating tube will be the same, regardless of whether half of its length is dipped into water or not.
When a tube is open at both ends, it can resonate at certain frequencies, known as its harmonics. The fundamental frequency, also known as the first harmonic, is the lowest frequency at which the tube can resonate. This frequency is determined by the length of the tube.
Effect of Water on the Length of the Tube:
When half of the length of the tube is dipped into water, the effective length of the tube changes. The water acts as a barrier, effectively closing one end of the tube. This means that the tube is no longer open at both ends, but rather open at one end and closed at the other.
Change in the Fundamental Frequency:
The fundamental frequency of the open tube is given by the equation:
f = v / (2L)
Where f is the frequency, v is the speed of sound, and L is the length of the tube. When the tube is dipped into water, the effective length of the tube changes to half of its original length. Let's denote this new length as L'.
The new fundamental frequency can be calculated using the equation:
f' = v / (4L')
Since L' is half of the original length, we can substitute L/2 for L' in the equation:
f' = v / (4 * L/2)
= v / (4 * L/2)
= v / (2L)
Conclusion:
From the equation above, we can see that the fundamental frequency of the tube remains unchanged when half of its length is dipped into water. This is because the effective length of the tube is halved, but the relationship between the frequency and length remains the same.
Therefore, the fundamental frequency of the open resonating tube will be the same, regardless of whether half of its length is dipped into water or not.
Community Answer
An open resonating tube has fundamental frequency of n when half of it...
Fundamental frequency for open organ pipe is given that is n where n =V/2l where l is the length of the organ pipe and V is the velocity of sound in air
now half of its length is dipped into water so it will become a closed organ pipe with l'=l/2
fundamental frequency of the closed organ pipe say n'= V/4l/2 that is V/2l
taking ratio
n/n'=V/2l/V/2l = 1
This implies that n/n'=1
so n=n'
hence fundamental frequency of open organ pipe after it is dipped half in water will be n
now half of its length is dipped into water so it will become a closed organ pipe with l'=l/2
fundamental frequency of the closed organ pipe say n'= V/4l/2 that is V/2l
taking ratio
n/n'=V/2l/V/2l = 1
This implies that n/n'=1
so n=n'
hence fundamental frequency of open organ pipe after it is dipped half in water will be n
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An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be?
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An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be?.
An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for An open resonating tube has fundamental frequency of n when half of its length is dipped into water then it's fundamental frequency will be?.
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