Hooke's law
Hooke's law
- The tension in an elastic string/spring is proportional to its extension $x$:
$$T = \frac{\lambda x}{L}$$
- where $L$ is the natural length and $\lambda$ the modulus of elasticity.
- Stretching stores elastic potential energy:
$$E = \frac{\lambda x^2}{2L}$$
Practice
An elastic string of natural length 2 m and modulus 50 N is stretched by 0.5 m. Find the tension T = λx/L (N).
T = λx/L = 50 × 0.5 / 2 = 25/2 = 12.5 N.
Practice
For the same string (λ = 50, x = 0.5, L = 2), find the stored elastic energy E = λx²/(2L) (J).
E = 50 × 0.5² / (2×2) = 50 × 0.25 / 4 = 12.5/4 = 3.125 J.
Practice
By Hooke's law, the tension is proportional to the extension.
T = λx/L, so doubling the extension doubles the tension.
You've got it
Key idea
- Hooke's law: $T = \dfrac{\lambda x}{L}$ (tension ∝ extension)
- stored elastic PE: $E = \dfrac{\lambda x^2}{2L}$
- $\lambda$ = modulus of elasticity, $L$ = natural length