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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) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01m
- d)The fundamental frequency is 100 Hz.
Correct answer is option 'B,C'. Can you explain this answer?
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Assuming the equation represents a standing wave on the string, we can determine some properties of the wave.
The equation y(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-1)t] can be broken down into two parts: the spatial part and the temporal part.
The spatial part is given by sin [(62.8m-1)x], which represents the position of the string at a given point x. The frequency of the spatial part is given by the coefficient in front of x, which is 62.8 m^-1. The wavelength of the wave can be calculated using the formula λ = 2π/k, where k is the wave number. In this case, k = 62.8 m^-1, so the wavelength is λ = 2π/(62.8 m^-1) = 0.1 m.
The temporal part is given by cos[(628s-1)t], which represents the time dependence of the wave. The frequency of the temporal part is given by the coefficient in front of t, which is 628 s^-1. The period of the wave can be calculated using the formula T = 2π/ω, where ω is the angular frequency. In this case, ω = 628 s^-1, so the period is T = 2π/(628 s^-1) = 0.01 s.
Since the wave is in its fifth harmonic, the wavelength of the wave is equal to 5 times the fundamental wavelength (λ_1), and the period of the wave is equal to 5 times the fundamental period (T_1).
Therefore, the fundamental wavelength is λ_1 = λ/5 = 0.1 m/5 = 0.02 m, and the fundamental period is T_1 = T/5 = 0.01 s/5 = 0.002 s.
In summary, the wave has a wavelength of 0.1 m, a period of 0.01 s, a fundamental wavelength of 0.02 m, and a fundamental period of 0.002 s.
The equation y(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-1)t] can be broken down into two parts: the spatial part and the temporal part.
The spatial part is given by sin [(62.8m-1)x], which represents the position of the string at a given point x. The frequency of the spatial part is given by the coefficient in front of x, which is 62.8 m^-1. The wavelength of the wave can be calculated using the formula λ = 2π/k, where k is the wave number. In this case, k = 62.8 m^-1, so the wavelength is λ = 2π/(62.8 m^-1) = 0.1 m.
The temporal part is given by cos[(628s-1)t], which represents the time dependence of the wave. The frequency of the temporal part is given by the coefficient in front of t, which is 628 s^-1. The period of the wave can be calculated using the formula T = 2π/ω, where ω is the angular frequency. In this case, ω = 628 s^-1, so the period is T = 2π/(628 s^-1) = 0.01 s.
Since the wave is in its fifth harmonic, the wavelength of the wave is equal to 5 times the fundamental wavelength (λ_1), and the period of the wave is equal to 5 times the fundamental period (T_1).
Therefore, the fundamental wavelength is λ_1 = λ/5 = 0.1 m/5 = 0.02 m, and the fundamental period is T_1 = T/5 = 0.01 s/5 = 0.002 s.
In summary, the wave has a wavelength of 0.1 m, a period of 0.01 s, a fundamental wavelength of 0.02 m, and a fundamental period of 0.002 s.
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A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equationy(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01md)The fundamental frequency is 100 Hz.Correct answer is option 'B,C'. Can you explain this answer?
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A horizontal stretched string fixed at two ends, is vibrating in its fifth harmonic according to the equationy(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01md)The fundamental frequency is 100 Hz.Correct 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 equationy(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01md)The fundamental frequency is 100 Hz.Correct 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 equationy(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01md)The fundamental frequency is 100 Hz.Correct 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 equationy(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01md)The fundamental frequency is 100 Hz.Correct 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 equationy(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01md)The fundamental frequency is 100 Hz.Correct 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 equationy(x, t) = 0.01m sin [(62.8m-1)x] cos[(628s-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 0.01md)The fundamental frequency is 100 Hz.Correct answer is option 'B,C'. Can you explain this answer?.
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