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Chapter 7: Problem 976

Cross-sectional area o if wire of length \(L\) is \(A\). Young's modulus of material is \(\mathrm{Y}\). If this wire acts as a spring what is the value of force constant? (A) (YA/L) (B) (YA/2L) (C) (2YA/L) (D) ( \(\mathrm{YL} / \mathrm{A})\)

Short Answer

Expert verified
The correct value of the force constant is (A) (YA/L).

Step by step solution

01

Recall Hooke's Law

Hooke's Law states that the force acting on a spring is proportional to the displacement of the spring from its equilibrium position. Mathematically, it is given by: \(F = -kx\) Where F is the force, k is the spring constant (or force constant), and x is the displacement of the spring from its equilibrium position.
02

Relate stress and strain using Young's modulus

We have to relate the force F to the length L and cross-sectional area A of the wire. To do this, we use the definition of Young's modulus, which states the relationship between stress and strain: \(Y = \dfrac{\text{stress}}{\text{strain}}\) Stress is defined as the force per unit area, while strain is defined as the change in length per unit length. We can write these definitions mathematically as: \(\text{stress} = \dfrac{F}{A}\) and \(\text{strain} = \dfrac{\Delta L}{L}\)
03

Find the force constant using Hooke's Law and Young's modulus

Now, we can substitute the expressions for stress and strain in the Young's modulus equation: \(Y = \dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}\) Next, we solve this equation for the force F: \(F = \dfrac{YA}{L}\Delta L\) By comparing this equation with Hooke's Law, we can see that the force constant, k, is given by: \(k = \dfrac{YA}{L}\)
04

Find the value of the force constant in the given options

Now, we can find the force constant value in the given options: (A) (YA/L) (B) (YA/2L) (C) (2YA/L) (D) (YL/A) We derived the force constant, k, as (YA/L), which matches option (A). Therefore, the correct answer is: (A) (YA/L)

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