Question

Derive Hooke's Law on Force constant​

Answer 1

Answer:

Hooke's law, F = kx, where the applied force F equals a constant k times the displacement or change in length x

Answer 2

Consider just a pair of atoms. For slight positional deviations E(0) (corresponding to the small-strain assumption of linear elasticity) around the smooth energy minimum E, regardless of its true functional form, can be expanded using a Taylor series,where the prime notation denotes derivatives with respect to position.

$\tt \: E(δ)=E(0)+E ′(0)δ+ \frac{E ′′(0)δ {}^{2} }{2!} + \frac{E ′′′(0)δ {}^{3} }{3!} + . \: . \: . \: . \: .$

Now let's set our energy reference to E(0)=0, note that E ′(0)=0. which describes the energy of an idealized spring with spring constant

$\tt \: E(δ)≈ \frac{1}{2} E ′′(0)δ {}^{2} ,$

The derivative of this equation with respect to position provides the restoring force, which is

$\tt \: F(δ)≈E ′′(0)δ.$

Now define stress as σ=F/A and engineering strain as ϵ=δ/L nd we have

$\tt \: σ= \frac{LE ′′(0) A}{ϵ} \\ \tt \: \: σ=Yϵ,$

which is Hooke's Law for a corresponding elastic modulus.

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