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The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.?
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The vibration of a string of length 20 cm and mass 2 g stretched betwe...
Given information:
- Length of the string, L = 20 cm
- Mass of the string, m = 2 g
- Length of the closed end tube, L' = 40 cm
- Difference in frequencies between the tube in its fundamental mode and the string in its first overtone, Δf = 10 Hz
To find:
- The tension in the string
Assumptions:
- The string and the tube are made of the same material, so the speed of sound in both is the same.
- The vibrating string and the air column in the tube are in resonance.
Explanation:
1. Understanding the fundamental frequency and overtone:
- The fundamental frequency is the lowest possible frequency at which a system can vibrate.
- The first overtone is the second harmonic, which is the second lowest frequency at which a system can vibrate.
2. Relationship between frequency and wavelength:
- The frequency of a wave is inversely proportional to its wavelength, given by the equation: f = v/λ
- Where f is the frequency, v is the velocity (speed of sound in this case), and λ is the wavelength.
3. Relationship between wavelength and length:
- For a vibrating string, the wavelength of the fundamental mode is twice the length of the string (λ = 2L).
- For a closed end tube, the wavelength of the fundamental mode is four times the length of the tube (λ = 4L').
4. Relationship between frequency and tension:
- The frequency of a vibrating string is directly proportional to the square root of the tension in the string (f ∝ √T).
- As the tension in the string decreases, the frequency decreases.
5. Calculating the frequencies:
- Let's assume the frequency of the tube in its fundamental mode is f1 and the frequency of the string in its first overtone is f2.
- Using the relationships mentioned above, we can write the following equations:
f1 = v/4L' (1)
f2 = (2v)/2L (2)
6. Calculating the tension in the string:
- From equation (1), we can rewrite it as:
v = 4L'f1 (3)
- From equation (2), we can rewrite it as:
v = Lf2 (4)
7. Equating equations (3) and (4):
4L'f1 = Lf2
f1 = f2/4L'/L
f1 = f2/8
8. Calculating the difference in frequencies:
- Given that the difference in frequencies is 10 Hz, we can write:
Δf = f1 - f2 = f2/8 - f2 = -7f2/8
9. Finding the tension in the string:
- Given that the difference in frequencies decreases as the tension in the string decreases, we can write:
Δf = -7f2/8 = 10 Hz
f2 = -80/7 Hz
- Using equation (2), we can solve for the tension in the string:
-
- Length of the string, L = 20 cm
- Mass of the string, m = 2 g
- Length of the closed end tube, L' = 40 cm
- Difference in frequencies between the tube in its fundamental mode and the string in its first overtone, Δf = 10 Hz
To find:
- The tension in the string
Assumptions:
- The string and the tube are made of the same material, so the speed of sound in both is the same.
- The vibrating string and the air column in the tube are in resonance.
Explanation:
1. Understanding the fundamental frequency and overtone:
- The fundamental frequency is the lowest possible frequency at which a system can vibrate.
- The first overtone is the second harmonic, which is the second lowest frequency at which a system can vibrate.
2. Relationship between frequency and wavelength:
- The frequency of a wave is inversely proportional to its wavelength, given by the equation: f = v/λ
- Where f is the frequency, v is the velocity (speed of sound in this case), and λ is the wavelength.
3. Relationship between wavelength and length:
- For a vibrating string, the wavelength of the fundamental mode is twice the length of the string (λ = 2L).
- For a closed end tube, the wavelength of the fundamental mode is four times the length of the tube (λ = 4L').
4. Relationship between frequency and tension:
- The frequency of a vibrating string is directly proportional to the square root of the tension in the string (f ∝ √T).
- As the tension in the string decreases, the frequency decreases.
5. Calculating the frequencies:
- Let's assume the frequency of the tube in its fundamental mode is f1 and the frequency of the string in its first overtone is f2.
- Using the relationships mentioned above, we can write the following equations:
f1 = v/4L' (1)
f2 = (2v)/2L (2)
6. Calculating the tension in the string:
- From equation (1), we can rewrite it as:
v = 4L'f1 (3)
- From equation (2), we can rewrite it as:
v = Lf2 (4)
7. Equating equations (3) and (4):
4L'f1 = Lf2
f1 = f2/4L'/L
f1 = f2/8
8. Calculating the difference in frequencies:
- Given that the difference in frequencies is 10 Hz, we can write:
Δf = f1 - f2 = f2/8 - f2 = -7f2/8
9. Finding the tension in the string:
- Given that the difference in frequencies decreases as the tension in the string decreases, we can write:
Δf = -7f2/8 = 10 Hz
f2 = -80/7 Hz
- Using equation (2), we can solve for the tension in the string:
-
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The vibration of a string of length 20 cm and mass 2 g stretched betwe...
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The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.?
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The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.?.
The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.?.
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The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.?, a detailed solution for The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.? has been provided alongside types of The vibration of a string of length 20 cm and mass 2 g stretched between two ends and those in a closed end tube of length 40 cm are compared. The difference in the frequencies of the tube in its fundamental mode and the string in its first overtone is 10 Hz. On decreasing the tension in the string , the difference in their frequencies decreases , then find the tension in the string.? theory, EduRev gives you an
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