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Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]?
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Calculate the natural frequency of the iron rod of 0.03m length. The d...
Calculation of Natural Frequency of Iron Rod
Given:
Length of the iron rod (l) = 0.03m
Density of iron (ρ) = 23230 kg/m^3
Young's modulus (E) = 1.6x10^10 N/m^2
Formula:
The natural frequency (f) of a rod is given by the formula:
f = (1/2π) * √(E/ρ) * (A/l)^2
Where,
A = Cross-sectional area of the rod
Calculation:
Cross-sectional area of the rod (A) = πr^2
Given, Length of the rod (l) = 0.03m
So, radius (r) = diameter/2 = 0.01m/2 = 0.005m
Therefore, A = π(0.005)^2 = 7.85x10^-5 m^2
Substituting the given values in the formula, we get:
f = (1/2π) * √(E/ρ) * (A/l)^2
f = (1/2π) * √(1.6x10^10 / 23230) * (7.85x10^-5 / 0.03)^2
f = 37.24 kHz (approx)
Answer:
The natural frequency of the iron rod of 0.03m length is 37.24 kHz.
Given:
Length of the iron rod (l) = 0.03m
Density of iron (ρ) = 23230 kg/m^3
Young's modulus (E) = 1.6x10^10 N/m^2
Formula:
The natural frequency (f) of a rod is given by the formula:
f = (1/2π) * √(E/ρ) * (A/l)^2
Where,
A = Cross-sectional area of the rod
Calculation:
Cross-sectional area of the rod (A) = πr^2
Given, Length of the rod (l) = 0.03m
So, radius (r) = diameter/2 = 0.01m/2 = 0.005m
Therefore, A = π(0.005)^2 = 7.85x10^-5 m^2
Substituting the given values in the formula, we get:
f = (1/2π) * √(E/ρ) * (A/l)^2
f = (1/2π) * √(1.6x10^10 / 23230) * (7.85x10^-5 / 0.03)^2
f = 37.24 kHz (approx)
Answer:
The natural frequency of the iron rod of 0.03m length is 37.24 kHz.
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Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]?
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Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]?.
Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Calculate the natural frequency of the iron rod of 0.03m length. The density of iron is 23230 kg/m^3 and Young's modulus is 1 1.6x1010 N/m^2. [Ans. 37.24 kHz]?.
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