Equilibrium Series · Part 3

HOOKE'S
LAW

You've just landed — and your car's suspension just compressed by a few centimetres. That's Hooke's Law. A trampoline launches you upward — Hooke's Law again. Your pen click, your door hinge, your heartbeat sensor — all springs, all Hooke's Law.

In Parts 1 and 2, you mastered equilibrium of forces. Now we go deeper: what happens to a spring when a force is applied? How does tension relate to stretch? Today you derive the answer yourself.

E2 · Explore

The Spring Experiment

Three springs — soft, medium, and stiff. For each one, drag the weight slider to apply a force. Watch the spring extend. Record the data. See the pattern emerge.

APPLIED FORCE (N): 0 N

0 N10 N20 N30 N40 N

Tension T

0 N

Δx (elongation)

0.00 m

📊 Data Table — Soft Spring

#	Force T (N)	Elongation Δx (m)	T / Δx
Apply a force and click Record

📈 T vs Δx Graph (live)

Observation Question · After Experimenting

After recording data for at least 3 force values, what pattern do you notice in the T / Δx column of your data table?

AThe ratio T / Δx keeps changing — no clear pattern

BThe ratio T / Δx is approximately constant for a given spring

CT / Δx increases as force increases

DT / Δx decreases as elongation increases

E3 · Explain

Deriving Hooke's Law

Your data showed a constant ratio T/Δx. Let's formalise this discovery step by step and derive the complete law — including what the graph tells us.

Step 1 · The Observation

From your experiment: as the applied force T (tension in the spring) increases, the elongation Δx increases proportionally. Double the force → double the elongation.

T ∝ Δx

Step 2 · Introduce the Constant

To convert proportionality into an equation, we introduce a constant k — the spring constant (or force constant). It measures how stiff the spring is.

T = k · Δx

Hooke's Law

Tension / Restoring force (N)

Spring constant (N/m)

Δx

Elongation or compression (m)

Step 3 · The Graph

Plot T (vertical axis) against Δx (horizontal axis).

✓ Straight line through the origin
✓ Positive gradient
✓ Gradient = k (spring constant)
✓ Greater k → steeper line → stiffer spring

gradient = ΔT / Δ(Δx) = k

T vs Δx — three springs compared

Step 4 · The Elastic Limit ⚠

Hooke's Law only holds within the elastic limit. If you stretch a spring beyond this point, the spring is permanently deformed — it does not return to its original length. The T vs Δx graph becomes curved beyond this point. The spring has left its elastic region and entered its plastic region.

Elastic region

Spring returns to original length · T = k·Δx applies · Graph is linear

Plastic region

Permanent deformation · Hooke's Law breaks down · Graph curves

Step 5 · Sign Convention (Compression vs Elongation)

Hooke's Law also applies when a spring is compressed. The restoring force always acts opposite to the displacement: if you stretch the spring downward (Δx positive), the tension pulls upward (T positive). If you compress upward (Δx negative), the restoring force pushes downward (T negative). This is the restoring force principle.

T = k · Δx works for both elongation (+Δx) and compression (−Δx)

🧮 Interactive: Read k from a Graph

The graph below shows T vs Δx for an unknown spring. Two points on the line are marked. Drag the slider to pick your two points, then calculate k = ΔT / Δ(Δx).

Click two points on the graph to calculate k

From the graph above, what is the spring constant k?

Use the two marked points to calculate gradient = ΔT / Δ(Δx)

Ak = 25 N/m

Bk = 50 N/m

Ck = 100 N/m

Dk = 200 N/m

E4 · Elaborate

Hooke's Law in the Real World

Springs are everywhere. Every one of these follows T = k·Δx — within the elastic limit.

🚗

Car Suspension

Each wheel sits on a spring. When you hit a pothole, the spring compresses (−Δx), absorbing the shock so passengers don't feel it. Lebanon's roads make this very relevant!

T = k · Δx → k must be chosen for car's mass

🏋️

Weighing Scale

A spring scale works exactly on Hooke's Law. The elongation Δx is proportional to weight (force). The scale just reads Δx and converts it to mass. Every market stall uses this.

W = k · Δx → m = kΔx / g

🏹

Archery Bow

A bow stores elastic potential energy as you pull the string back (elongation Δx). The stiffer the bow (higher k), the more force required — and the more energy released when fired.

E_elastic = ½ k (Δx)²

❤️

Medical Stents

Tiny metal springs (stents) are inserted into blocked arteries. Their spring constant is designed so they hold arteries open with exactly the right force — not crushing the vessel.

Precise k selection = life-saving medicine

🎸

Guitar Strings

A guitar string behaves like a spring under tension. Tightening the tuning peg increases the spring tension T, which raises the vibration frequency and the musical pitch.

Higher T → higher frequency → higher note

🌉

Suspension Bridges

The main cables of suspension bridges stretch elastically under load. Engineers calculate Δx under maximum traffic load to ensure the bridge stays within safe elastic limits.

k must keep Δx within safe range

🔄 Elongation vs Compression — Interactive

Drag the slider left (compression) or right (elongation). Observe how the spring responds and how T = k·Δx applies in both directions. The restoring force always opposes displacement.

← COMPRESS EXTEND →

Δx = 0 m | T = 0 N | Direction: at rest

E4 · Evaluation

Evaluation Questions

Five questions covering all levels — from direct recall to multi-step calculation and analysis.

Question 1 · Definition

Hooke's Law states that, within the elastic limit, the tension in a spring is:

AInversely proportional to the elongation Δx

BDirectly proportional to the elongation Δx

CEqual to the square of the elongation Δx

DIndependent of the elongation Δx

Question 2 · Calculation

A spring has a spring constant k = 60 N/m. A force of 18 N is applied. What is the elongation Δx?

AΔx = 1080 m

BΔx = 3.33 m

CΔx = 0.30 m

DΔx = 0.03 m

Question 3 · Find k from Data

A spring extends by Δx = 5 cm when a force of T = 4 N is applied. What is the spring constant k?

Ak = 0.20 N/m

Bk = 20 N/m

Ck = 80 N/m

Dk = 400 N/m

Question 4 · Graph Analysis

A T vs Δx graph for Spring A has gradient = 120 N/m. Spring B has gradient = 45 N/m. Which statement is correct?

ASpring B is stiffer because it needs more force to extend the same distance

BSpring A is stiffer — higher gradient = higher k = more force needed per metre of elongation

CBoth springs are the same stiffness

DThe gradient does not represent the spring constant

Question 5 · Critical Thinking

A student stretches a spring beyond its elastic limit. She then removes the force. What happens, and why does Hooke's Law no longer apply?

AThe spring returns to normal — Hooke's Law always applies for springs

BThe spring is permanently deformed — internal structure changed, so T ≠ k·Δx any more

CThe spring becomes stronger — stretching increases its spring constant

DThe spring breaks immediately upon passing the elastic limit

Question 6 · Written Response

Explain how you would use a spring and a set of known masses to determine the spring constant k experimentally. Include: what you measure, how you use T = k·Δx, and what graph you would draw.

Written response saved with your submission and reviewed by Ahmad.

E5 · Submit

Submit Your Lab Report

Choose your section and name, then submit everything to Ahmad's grade sheet.

Section

Your Name

0 / 7 completed

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Lab Score

Ahmad reviews your data table, graph readings, and written response in class.