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We know that when a spring is stretched from one end to another, it tries to come back to its original orientation. It happens because there is an interatomic force of attraction between the molecules tightly packed in the solid material like a spring.

Now, when we exert force on these molecules to bring them away from their lattice points, they recover their position and spring regains its shape.

Now, if the same helical spring is extended by a load, we can find the value of spring constant of helical spring by performing the helical spring experiment and this article discusses the same.

A helical spring is the most commonly used mechanical spring in which a wire is wounded in a coil that seems like a screw thread. It is designed to carry, pull, or push loads.

We can find the usage of the twisted helical or torsion springs in engine starters and hinges.

Now, let’s study how we can use the helical spring to do a spring constant experiment.

Our objective is to find the force constant of the helical spring by plotting a graph between load and extension.

A stiff support

A spring

Six loads of the range 20 to 50 grams

A 30 or 50 g hanger

A movable sharp pointer

A wooden scale

A hook

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If F is the force applied, x is the displacement of the spring, whose magnitude equals the magnitude of the force applied. The equation form is:

F ∝ x

Here,

F = - k x

This means ‘k’ is the force constant and the regaining is specified with a negative sign. A force constant or the spring constant ‘k’ has no unit.

Basically, spring has the formula mentioned in equation (1). Since we are here to find the force constant of helical spring by plotting a graph between load and extension, so instead of using the displacement term, we will use a length by which a helical spring got an extension under the influence of the load.

Here, we have the following equation:

F = kL

L = extension in the length of the helical spring, which is positive by nature.

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Suspend the helical spring from a rigid support and attach the pointer vertically and the hook from its lower free end. Hang a 40 g hanger from the hook.

Arrange the vertical wooden scale in such a way that the tip of the pointer comes over the divisions on the scale and can slide over it accordingly without touching it.

Firstly, add a dead weight to the hanger by keeping the spring vertical. Jot down the reading on the scale and record it in the loading column against the zero loads.

The weights of the range 20 g to 50 g are added one by one to the slotted weight to the hanger till the maximum load is reached. In each case, when the tip of the pointer moves down, note the reading of the pointer.

Then, each weight is removed one by one and the reading of the pointer is noted in each case of unloading.

Now, wait for some time till the pointer comes to rest. Repeat step 3.

As the pointer moves up. Repeat step 5 and 6 and record the reading in the unloading column.

Repeat step 7 till the only hanger is free.

Record your observation and perform the helical spring experiment calculations.

The average of the readings for each load/weight during the loading and unloading process is calculated in each case. Let m0, m1,m2, m3…etc.., be the average readings of the pointer for the loads w0, (w0 + 50),

(w0+ 100), (w0+ 45), etc.

From this, extension, ‘L’ (in m) for the loads (w0 + 50), (w0+ 100), (w0+ 150), etc. , are calculated as (m1 - m0), (m2 - m0 ), (m3 - m0), respectively.

In each case, k = mg/l is calculated.

Here, g is the acceleration due to gravity = 9.8 ms-2.The average value of k gives the helical spring constant in N/m.

Now, after making the helical spring experiment calculations, record these in the following table:

After performing all the experimentations, we get the following graph:

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OX = ——— kg wt

OY = ——— m

Value of spring constant of helical spring ‘K’ = ——— Nm⁻¹

By calculation, the value of the spring constant of helical spring is = ………….N/m.

From the helical spring load-extension graph, the force constant of the helical spring =……….N/m.

FAQ (Frequently Asked Questions)

Q1: Define a Rigid Body.

Ans: A rigid body is said to have a fixed length, shape, and volume. In rotational dynamics, we call a rigid body a perfect body because all the particles residing in the solid remain at a fixed position from their axis of rotation before and after rotation.

Practically, no rigid body exists. It is all we consider in the rotational dynamics, as this topic focuses on the intuition and imagination of the person.

Q2: Define Spring Constant.

Ans: Spring constant is the force applied if the displacement ‘x’ in the spring is unity or one.

If a force F applied is considered that extends the spring so that it displaces the equilibrium position by ‘x’ metres, we express ‘k’ in Newton per meter (N/m).

For example, A spring with a load of 10 Kg is stretched by 30 cm. The spring constant comes in the following manner:

F = ma = 10 x 0.3 = 3 N

k = - F/x

So,

k = - 3/0.3 = - 10 N/m

Q3: A Girl Weighing 30 Pounds Stretches a Spring by 40 cm. Calculate the Spring Constant of the Spring.

Solution:

Given m = 30 lbs

=> 30 / 2.2 = 13.64 Kg

Displacement x = 40 cm

The force F = ma = 13.64 × 9.8 = 133.672 N

The spring constant formula is given by:

k = − F/x

= – 133.672 / 0.4 = – 334.18 N/m.