Hooke’s Law (AQA A-Level Further Maths): Revision Notes

Worked Example: Single String Calculations

Problem: A string with natural length 5 m and modulus of elasticity 10 N is extended 0.4 m.

a) What is the stiffness of the string?

The stiffness is:

$k = \frac{\lambda}{l} = \frac{10}{5} = 2 \text{ Nm}^{-1}$

b) Calculate the tension in the string and the EPE stored in it.

Using Hooke's law:

$T = \frac{\lambda}{l}x = \frac{10}{5} \times 0.4 = 0.8 \text{ N}$

The elastic potential energy stored is:

$\text{EPE} = \frac{\lambda}{2l}x^2 = \frac{10}{2 \times 5} \times 0.4^2 = 0.16 \text{ J}$

c) If the string is extended a further 0.3 m, how much extra EPE is stored in it?

The increase in EPE is:

$\Delta E = \frac{\lambda}{2l}(x_2^2 - x_1^2) = \frac{10}{2 \times 5}(0.7^2 - 0.4^2) = 0.33 \text{ J}$

Worked Example: Multiple Strings in Equilibrium

Problem: Two strings $OA$ and $OB$ are tied to an object $O$ which is held on a smooth table between two fixed points $A$ and $B$ which are 1.0 m apart.

• String $OA$ has natural length 0.4 m and modulus of elasticity 12 N
• String $OB$ has natural length 0.3 m and modulus of elasticity 18 N

Find the extensions of the strings and their total EPE.

Solution:

The object is in equilibrium, so the horizontal forces balance. This means the tensions in both strings are equal:

$T_1 = T_2$

Using Hooke's law for each string:

For $OA$: $T_1 = \frac{\lambda_1}{l_1}x_1 = \frac{12}{0.4}x_1 = 30x_1$

For $OB$: $T_2 = \frac{\lambda_2}{l_2}x_2 = \frac{18}{0.3}x_2 = 60x_2$

Since $T_1 = T_2$:

$30x_1 = 60x_2$

Therefore:

$x_1 = 2x_2 \quad \text{(equation 1)}$

The total distance between $A$ and $B$ is:

$\text{Distance } AB = 1.0 = 0.4 + 0.3 + x_1 + x_2$

Simplifying:

$0.3 = x_1 + x_2 \quad \text{(equation 2)}$

Solving the simultaneous equations by substituting equation 1 into equation 2:

$0.3 = 2x_2 + x_2 = 3x_2$

$x_2 = 0.1 \text{ m}$

$x_1 = 2 \times 0.1 = 0.2 \text{ m}$

The total EPE is:

$\text{Total EPE} = \frac{\lambda_1}{2l_1}x_1^2 + \frac{\lambda_2}{2l_2}x_2^2$

$= \frac{12}{2 \times 0.4} \times 0.2^2 + \frac{18}{2 \times 0.3} \times 0.1^2$

$= 0.6 + 0.3 = 0.9 \text{ J}$

Worked Example: Energy Conservation with Vertical String

Problem: A vertical elastic string of natural length $AB = 1$ m with $\lambda = 49$ N has end $A$ fixed to a ceiling and end $B$ fixed to a mass of 2 kg. Take $g = 9.8$ m s$^{-2}$.

a) The mass is lowered gently until in equilibrium at point $O$. Calculate the extension when at $O$.

b) It is then pulled down a further 20 cm to point $P$ and released. Calculate its velocity $v$ when passing $O$ and the greatest height that it reaches. Describe its subsequent motion.

Solution:

Part a: When in equilibrium at $O$ with extension $x_{eq}$:

The tension in the string equals the weight of the 2 kg mass:

$T = mg$

Using Hooke's law:

$\frac{\lambda}{l}x_{eq} = mg$

$\frac{49}{1} \times x_{eq} = 2 \times 9.8$

$x_{eq} = \frac{19.6}{49} = 0.4 \text{ metres}$

Part b: Energy equation from $P$ to $O$:

At point $P$, the mass is at rest (KE = 0) and the extension is 0.6 m. At point $O$, the extension is 0.4 m and the mass has velocity $v$.

Using energy conservation (taking $O$ as the reference level for GPE):

$\text{Gain in KE} + \text{Gain in GPE} = \text{Loss of EPE}$

$\frac{1}{2} \times 2 \times v^2 + 2 \times g \times 0.20 = \frac{49}{2 \times 1}(0.60^2 - 0.40^2)$

$v^2 + 2 \times 9.8 \times 0.20 = \frac{49}{2}(0.36 - 0.16)$

$v^2 = 4.9 - 3.92 = 0.98$

$v = 0.99 \text{ m s}^{-1}$

Greatest height: At the greatest height, the velocity is zero, so KE = 0.

Let the extension at the highest point be $x$. Using energy conservation from $P$ to the highest point (taking $P$ as reference):

$\text{Loss of EPE} = \text{Gain in GPE}$

$\frac{49}{2 \times 1}(0.6^2 - x^2) = 2 \times g \times (0.6 - x)$

$\frac{49}{2}(0.6 - x)(0.6 + x) = 2 \times 9.8 \times (0.6 - x)$

Dividing both sides by $(0.6 - x)$:

$\frac{49}{2}(0.6 + x) = 19.6$

$0.6 + x = \frac{19.6 \times 2}{49} = 0.8$

$x = 0.2 \text{ m}$

Alternatively, if $0.6 - x = 0$, then $x = 0.6$ m (this is point $P$ itself).

When $x = 0.6$, the mass is at $P$. When $x = 0.2$, the mass is at the highest point.

The greatest height is $1.0 + 0.2 = 1.2$ metres below $A$.

Subsequent motion: The midpoint of the oscillation is at point $O$ (the equilibrium position), so the mass oscillates between $P$ and the highest point, moving up and down through $O$.