Elastic potential energy of spring — Hooke's law applications and combinations

Question

How do we calculate the elastic potential energy stored in a spring? How does Hooke’s law apply to series and parallel spring combinations?

(CBSE 11, JEE Main, NEET — spring energy and spring combinations appear frequently in both mechanics and SHM chapters)

Solution — Step by Step

F = -kx

where $k$ = spring constant (N/m) and $x$ = displacement from natural length. The negative sign indicates a restoring force (opposing displacement).

The spring constant $k$ measures stiffness — higher $k$ means a stiffer spring that requires more force for the same extension. $k$ depends on the material, coil diameter, wire thickness, and number of turns.

The work done in stretching/compressing a spring from natural length by $x$ :

PE_{spring} = \frac{1}{2}kx^2

This energy is stored in the spring and is released when the spring returns to natural length.

Since $PE \propto x^2$ , doubling the extension stores 4 times the energy. This quadratic relationship is crucial — it means the last bit of stretching stores much more energy than the first bit.

Work done between two extensions $x_1$ and $x_2$ :

W = \frac{1}{2}k(x_2^2 - x_1^2)

Springs in series (end to end):

Same force through each spring, extensions add up

\frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2}

Series combination is SOFTER (smaller $k_{eq}$ )

Springs in parallel (side by side):

Same extension, forces add up

k_{eq} = k_1 + k_2

Parallel combination is STIFFER (larger $k_{eq}$ )

Note: this is the opposite pattern from resistors (where series adds and parallel uses reciprocals). For springs, series uses reciprocals and parallel adds directly.

Two springs ( $k_1 = 200$ N/m and $k_2 = 300$ N/m) are connected in series. The combination is compressed by 0.1 m. Find the total energy stored.

First, find $k_{eq}$ :

\frac{1}{k_{eq}} = \frac{1}{200} + \frac{1}{300} = \frac{3+2}{600} = \frac{5}{600}

k_{eq} = 120 \text{ N/m}

Energy stored:

PE = \frac{1}{2}k_{eq}x^2 = \frac{1}{2}(120)(0.1)^2 = \frac{1}{2}(120)(0.01) = \mathbf{0.6 \text{ J}}

flowchart TD
    A["Spring Problem"] --> B{"Single spring or combination?"}
    B -->|"Single"| C["Use F = kx and PE = ½kx²"]
    B -->|"Combination"| D{"How are springs connected?"}
    D -->|"Series (end to end)"| E["1/keq = 1/k₁ + 1/k₂<br/>Softer overall"]
    D -->|"Parallel (side by side)"| F["keq = k₁ + k₂<br/>Stiffer overall"]
    E --> G["Use keq in PE = ½keq·x²"]
    F --> G

Why This Works

Hooke’s law describes a linear restoring force, which stores energy as a quadratic function of displacement. The $\frac{1}{2}kx^2$ formula comes from integrating the force over displacement: $W = \int_0^x kx\, dx = \frac{1}{2}kx^2$ . The force increases linearly with $x$ , so the average force over the displacement is $\frac{1}{2}kx$ , and $W = F_{avg} \times x = \frac{1}{2}kx^2$ .

For combinations, the key principle is: in series, forces are equal (same tension throughout); in parallel, displacements are equal (both springs stretch by the same amount).

Common Mistake

Students apply $PE = \frac{1}{2}kx^2$ using the individual spring constant when the spring is part of a combination. In a series combination compressed by total distance $x$ , each spring compresses by different amounts ( $x_1 \neq x_2$ ). The total energy is $\frac{1}{2}k_{eq}x^2$ , NOT $\frac{1}{2}k_1 x^2 + \frac{1}{2}k_2 x^2$ . Use the equivalent spring constant with the total displacement.

Series springs have the SAME formula pattern as parallel resistors ( $1/k_{eq} = 1/k_1 + 1/k_2$ ). This is counterintuitive but easy to remember: springs and resistors have OPPOSITE combination rules. If you can do one, just swap the formulas for the other.

Question

Solution — Step by Step

Why This Works

Common Mistake

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