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A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)
- a)The number of nodes is 5
- b)The length of the string is 0.25 m
- c)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 m
- d)The fundamental frequency is 100 Hz
Correct answer is option 'B,C'. Can you explain this answer?
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Verified Answer
A horizontal stretched string, fixed at two ends, is vibrating in its ...
y = [0.01 sin (62.8x)] cos (628 t).
The midpoint M is an antinode and has the maximum displacement = 0.01 m
The fundamental frequency = 
Most Upvoted Answer
A horizontal stretched string, fixed at two ends, is vibrating in its ...
The Wave Equation Analysis
The given wave equation is y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. This represents a standing wave on a string fixed at both ends.
Understanding the Harmonics
- In a string fixed at both ends, the nth harmonic has n nodes. Thus, for the fifth harmonic:
- Number of Nodes: There are 5 nodes, but actually, there are 4 segments of the wave, creating 5 points including the ends where the displacement is zero.
Length of the String
- The wave equation indicates the wave number k = 62.8 m–1.
- The relationship between the wave number and the length of the string (L) is given by:
L = n * (λ/2), where n is the harmonic number and λ is the wavelength.
- The wavelength (λ) can be calculated as λ = 2π/k = 2π/62.8.
- For the fifth harmonic (n=5), the length of the string is:
L = (5/2) * (2π/k) = 5/62.8 = 0.25 m.
Maximum Displacement
- The amplitude of the wave is given as 0.01 m, which represents the maximum displacement from the equilibrium position.
- Therefore, the maximum displacement at the midpoint (which is a node) is indeed 0.01 m.
Fundamental Frequency
- The fundamental frequency (f1) can be derived from the relationship f1 = v/λ, where v is the wave speed.
- The wave speed can be calculated from the angular frequency ω = 628 s–1 and wave number k:
v = ω/k = 628/62.8 = 10 m/s.
- The wavelength for the fundamental mode (λ1) is 2L = 2 * 0.25 m = 0.5 m.
- Thus, the fundamental frequency f1 = v/λ1 = 10/0.5 = 20 Hz, not 100 Hz.
Conclusion
- Correct answers:
- B: The length of the string is 0.25 m.
- C: The maximum displacement is 0.01 m.
- Incorrect statements:
- A: The number of nodes is 5 (it should be 4 segments).
- D: The fundamental frequency is 100 Hz (it is actually 20 Hz).
The given wave equation is y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. This represents a standing wave on a string fixed at both ends.
Understanding the Harmonics
- In a string fixed at both ends, the nth harmonic has n nodes. Thus, for the fifth harmonic:
- Number of Nodes: There are 5 nodes, but actually, there are 4 segments of the wave, creating 5 points including the ends where the displacement is zero.
Length of the String
- The wave equation indicates the wave number k = 62.8 m–1.
- The relationship between the wave number and the length of the string (L) is given by:
L = n * (λ/2), where n is the harmonic number and λ is the wavelength.
- The wavelength (λ) can be calculated as λ = 2π/k = 2π/62.8.
- For the fifth harmonic (n=5), the length of the string is:
L = (5/2) * (2π/k) = 5/62.8 = 0.25 m.
Maximum Displacement
- The amplitude of the wave is given as 0.01 m, which represents the maximum displacement from the equilibrium position.
- Therefore, the maximum displacement at the midpoint (which is a node) is indeed 0.01 m.
Fundamental Frequency
- The fundamental frequency (f1) can be derived from the relationship f1 = v/λ, where v is the wave speed.
- The wave speed can be calculated from the angular frequency ω = 628 s–1 and wave number k:
v = ω/k = 628/62.8 = 10 m/s.
- The wavelength for the fundamental mode (λ1) is 2L = 2 * 0.25 m = 0.5 m.
- Thus, the fundamental frequency f1 = v/λ1 = 10/0.5 = 20 Hz, not 100 Hz.
Conclusion
- Correct answers:
- B: The length of the string is 0.25 m.
- C: The maximum displacement is 0.01 m.
- Incorrect statements:
- A: The number of nodes is 5 (it should be 4 segments).
- D: The fundamental frequency is 100 Hz (it is actually 20 Hz).
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A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. Can you explain this answer?
Question Description
A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. 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 horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. 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 horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. Can you explain this answer?.
A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. 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 horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. 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 horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. Can you explain this answer?.
Solutions for A horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. 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 horizontal stretched string, fixed at two ends, is vibrating in its fifth harmonic according to the equation, y(x, t) = (0.01 m) sin [(62.8 m–1)x] cos[(628 s–1)t]. Assuming π = 3.14, the correct statement(s) is (are)a)The number of nodes is 5b)The length of the string is 0.25 mc)The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01 md)The fundamental frequency is 100 HzCorrect answer is option 'B,C'. Can you explain this answer? tests, examples and also practice JEE tests.
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