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A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.
Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is
Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is
- a)344
- b)336
- c)117.3
- d)109.3
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
A vibrating string of certain length ℓ under a tension T resonat...
The frequency (v) produced by the air column is given by
∴ The frequency of vibrating string = 340. Since this string produces 4 beats/sec with a tuning fork of frequency n therefore n = 340 + 4 or n = 340 – 4. With increase in tension, the frequency produced by string increases. As the beats/sec decreases therefore n = 340 + 4 = 344 Hz.
Most Upvoted Answer
A vibrating string of certain length ℓ under a tension T resonat...
Given information:
- Length of air column, L = 75 cm = 0.75 m
- Velocity of sound in air, v = 340 m/s
- First overtone (third harmonic) of air column resonates with the vibrating string
- The string generates 4 beats per second when excited along with a tuning fork of frequency n
- When tension of the string is slightly increased, the number of beats reduces to 2 per second
Finding the frequency of the tuning fork:
Let the frequency of the vibrating string be f1. Then, the frequency of the first overtone (third harmonic) of the air column is 3f1.
### Finding the frequency of vibration of the air column
The frequency of vibration of the air column can be found using the formula:
f = (nv)/(2L)
where n is the mode of vibration (first overtone in this case), v is the velocity of sound in air, and L is the length of the air column.
Substituting the given values, we get:
3f1 = (n * 340)/(2 * 0.75)
Simplifying, we get:
f1 = (n * 170)/(3 * 0.75)
### Finding the frequency of the tuning fork
When the vibrating string and the tuning fork are sounded together, beats are produced due to the difference in frequency. Let the frequency of the tuning fork be f2.
When the number of beats reduces from 4 per second to 2 per second, it means that the difference in frequency between the tuning fork and the vibrating string has reduced by half.
### Finding the difference in frequency before and after increasing tension
Let the tension in the string be T. When the tension is increased slightly, the frequency of vibration of the string also increases slightly. Let the new frequency be f1'.
The difference in frequency before and after increasing tension can be found using the formula:
Δf = (1/2) * |f1 - f2| = (1/2) * |f1' - f2|
where | | represents the absolute value.
### Using the information to find the frequency of the tuning fork
From the given options, we can try to find a value of n that satisfies all the given conditions.
Let's try option A: n = 344
Substituting n = 344 in the formulas, we get:
- f1 = (344 * 170)/(3 * 0.75) = 27413.33 Hz
- The difference in frequency before and after increasing tension is: Δf = (1/2) * |27413.33 - 344| = 13534.67 Hz
- After increasing tension, let the frequency of the string be f1' = f1 + Δf = 40948 Hz
- Now, the difference in frequency between the tuning fork and the vibrating string is: |f1' - f2| = 680 Hz
- Half of the above difference is 340 Hz, which matches the given information that the number of beats reduces to 2 per second.
Therefore, the correct option is A: 344 Hz.
- Length of air column, L = 75 cm = 0.75 m
- Velocity of sound in air, v = 340 m/s
- First overtone (third harmonic) of air column resonates with the vibrating string
- The string generates 4 beats per second when excited along with a tuning fork of frequency n
- When tension of the string is slightly increased, the number of beats reduces to 2 per second
Finding the frequency of the tuning fork:
Let the frequency of the vibrating string be f1. Then, the frequency of the first overtone (third harmonic) of the air column is 3f1.
### Finding the frequency of vibration of the air column
The frequency of vibration of the air column can be found using the formula:
f = (nv)/(2L)
where n is the mode of vibration (first overtone in this case), v is the velocity of sound in air, and L is the length of the air column.
Substituting the given values, we get:
3f1 = (n * 340)/(2 * 0.75)
Simplifying, we get:
f1 = (n * 170)/(3 * 0.75)
### Finding the frequency of the tuning fork
When the vibrating string and the tuning fork are sounded together, beats are produced due to the difference in frequency. Let the frequency of the tuning fork be f2.
When the number of beats reduces from 4 per second to 2 per second, it means that the difference in frequency between the tuning fork and the vibrating string has reduced by half.
### Finding the difference in frequency before and after increasing tension
Let the tension in the string be T. When the tension is increased slightly, the frequency of vibration of the string also increases slightly. Let the new frequency be f1'.
The difference in frequency before and after increasing tension can be found using the formula:
Δf = (1/2) * |f1 - f2| = (1/2) * |f1' - f2|
where | | represents the absolute value.
### Using the information to find the frequency of the tuning fork
From the given options, we can try to find a value of n that satisfies all the given conditions.
Let's try option A: n = 344
Substituting n = 344 in the formulas, we get:
- f1 = (344 * 170)/(3 * 0.75) = 27413.33 Hz
- The difference in frequency before and after increasing tension is: Δf = (1/2) * |27413.33 - 344| = 13534.67 Hz
- After increasing tension, let the frequency of the string be f1' = f1 + Δf = 40948 Hz
- Now, the difference in frequency between the tuning fork and the vibrating string is: |f1' - f2| = 680 Hz
- Half of the above difference is 340 Hz, which matches the given information that the number of beats reduces to 2 per second.
Therefore, the correct option is A: 344 Hz.
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A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct answer is option 'A'. Can you explain this answer?
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A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct 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 vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct 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 vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct answer is option 'A'. Can you explain this answer?.
A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct 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 vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct 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 vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct answer is option 'A'. Can you explain this answer?.
Solutions for A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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ample number of questions to practice A vibrating string of certain length ℓ under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slightly increased the number of beats reduces 2 per second.Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz isa)344b)336c)117.3d)109.3Correct answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
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