Understanding Hooke’s Law: Experiment, Observations, and Explanation

To verify Hooke’s Law, you can conduct a simple experiment using a spring and weights and follow the steps below.

Apparatus

1. A spring (helical coil spring)
2. A support stand or clamp
3. A ruler or measuring tape
4. Weights (e.g., metal washers or masses)
5. A balance or scale (optional, for measuring weights)

Procedure:

set the apparatus as shown

One end of the spring is securely fixed to a support stand or clamp, ensuring that the spring hangs vertically without touching any surfaces.

The length of the spring is determined by reading the pointer position on the ruler fixed besides the spring as in the figure below. This will serve as the reference length 𝐿₀=_______________

We Start by attaching a small weight of like 50g (0.5N) to the free end of the spring as shown.

We record the weight of the added mass 𝑚 in kg. The weight should be small enough to avoid damaging the spring but large enough to cause noticeable stretching.

With the weight attached, we measure and record the new length of the spring and then subtract the initial length 𝐿₀ from the stretched length L to find the extension e

Record the extension found from e=L-𝐿₀

We increase the load and record the corresponding extension in each case so as there is a table showing extensions given by different forces on the spring. The table is similar to the one shown below.

A graph of force F against the extension per each force resulting to a straight line through the origin as shown below:

Notes:

• Ensure that the spring is not stretched beyond its elastic limit, as this can cause permanent deformation and invalidate the results.
• Repeat the experiment multiple times to ensure accuracy and reliability of the measurements.
• Use a balance or scale to measure weights accurately, and be cautious when handling heavy objects to prevent accidents.

Observations

Provided the stretching weight is not too much, the spring always return to it’s original length when weight is unloaded.A plot of stretching force against extension is a straight line through the origin showing that the ratio of force against extension is a constant.

Explanations

Extension of a spring is directly proportional to the stretching force F. The stretching force can be exceeded beyond a certain value that causes permanent stretching that does allow return of original length on unloading.

The graph of extension against stretching force is as shown.

OP represents the permanent stretching or permanent extension of the spring.

Point E is known as the elastic limit of the spring. Beyond the elastic limit , further extension causes permanent extension.

Hooke’s law

This is a law that was developed by Robert Hooke after he investigated principles behind stretching of materials under forces and concluded his findings in Hooke’s law that states that:

For a helical spring or other elastic material, the extension is directly proportional to the stretching force provided the elastic limit is not exceeded.

The Hooke’s law cab be represented with mathematical notations as:

Force F ∝ extension e;

From the relationship above, an equation is developed such that:

F = ke where k is a constant of proportionality which depends on the material making the spring. The constant k is usually referred to as the spring constant.

k is obtained from the plotting of F against e as the gradient of the graph as shown below

That is;

gradient = k = change in force/ change in extension

The spring constant is expressed in Nm^-1 as it’s SI units.

Remember that some work is done whenever a force moves some distance s. that is; work = Fs.

Therefore, work is done by force when spring extends by distance e. The total work done by the masses stretching the spring is the average the work done by individual mass and can be obtained as area under the graph of force F against extension e.

The area under the graph is definitely the area of a triangle which will be obtained as work done = (1/2)Fe where F is the force applied and e is the extension produced by the force.

but F = ke and so we can substitute F for ke so that we have :

Questions for practice

1. A mass of 100g is suspended from the lower end of a spring . If the spring extends by 100 mm and the elastic limit of the spring is not exceeded, what is the spring constant.

Answer: 10 Nm^-1

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2. A metal cube suspended freely from the end of a spring causes it to stretch by 5.0 cm. A 500 g mass suspended from the same spring stretches it by 2.0 cm. If the elastic limit is not exceeded:

(a) Find the weight of the metal cube (answer: 12.5 N)

(b) By what length with the spring stretch if a mass of 1.5 kg is attached to it’s end? (answer: 6.0 cm)

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