Hooke’s Law

Hooke's Law is a basic concept in physics that explains how materials react when stretched or squeezed. It says that the force needed to stretch or compress something is directly related to how much it changes in size or shape, but this only holds true as long as the material stays within its elastic limit.

1.0Types of Forces

1. Deforming Forces: An external force that causes a change in the length, volume, or shape of a body is known as a deforming force.
2. Restoring Forces: When an external force acts on any object then an internal resistance is produced in the material due to the intermolecular forces which are called restoring force.
3. Elasticity: Elasticity is that property of a material of a body by virtue of which it opposes any change in its shape or size when deforming forces are applied on it, and recover its original state as soon as the deforming force is removed.

2.0Types of Bodies

1. Rigid Body: A body is said to be rigid if the relative positions of its constituent particles remain unchanged when external deforming forces are applied to it. The nearest approach to a rigid body is diamond or carborundum.
2. Perfectly Elastic Body: A body which perfectly regains its original form on removing the external deforming force, is defined as a perfectly elastic body. Example : quartz.It is quite close to a perfect elastic body.
3. Plastic Body: A body which does not have the property of opposing the deforming forces, is known as a plastic body. All bodies which remain in the deformed state even after the removal of the deforming forces are known as plastic bodies. Example : clay, wax, putty.

3.0Stress

The restoring force acting per unit area of the deformed body is called stress.

$Stress=\frac{\text{Internal Restoring Force}}{\text{Area of Cross-Section}}$        

$Stress=\frac{F_{external}\text{(At Equilibrium)}}{A}$

$S.IUnit:N/m^2$

$\text{Dimensions}:M^1{L}^{-1}{T}^{-2}$

4.0Strain

It is defined as the ratio of the change in length to the material's original length.

$\text{Strain}=\frac{\text{Change in the dimension of the body}}{\text{Original Dimension of the body}}$

It is a unitless and dimensionless quantity.

5.0Statement of Hooke’s Law

According to this law within the elastic limit the stress produced in a body is directly proportional to the corresponding strain.

$Stress\propto Strain\Rightarrow Stress=E\times Strain$

$\text{ E=Coefficient of Elasticity or Modulus of Elasticity}$

$E=\frac{Stress}{Strain}$

6.0Graph of Hookes’ Law

The slope of the stress & strain graph gives a coefficient of elasticity.

(A) E depends on :

1. Nature of material

2. Impurities

3. Temperature

(B) E independent from:

1. Stress

2. Strain

7.0Experiment of Hooke’s Law

• When an external force is applied to a body, the shape of the body changes or deformation occurs.
• Restoring forces are developed within the body, with a magnitude equal to the applied force, which tend to oppose the deformation.
• On removing the applied force, the body regains its original shape, provided the deformation is within the elastic limit.
• For small changes in length or shape within the elastic limit, the magnitude of the elongation or extension is directly proportional to the applied force, as described by Hooke’s Law.

 Load-Extension Graph for a Helical Spring

Calculating Spring Constant

Slope of the straight line graph $Tan\theta =\frac{AB}{OB}=\frac{x}{F}$

Spring Constant, $k=\frac{F}{x}=\frac{1}{\text{Slope of the Graph}}$

Spring Constant,

$k=\frac{OB}{AB}=\frac{F_B-F_O}{x_A-x_B}$

where $x_A$ and $x_B$ are the corresponding extensions at points A and B (or O) respectively where $F_B$and $F_O$ are the loads (forces) at points B and O.

8.0Hooke’s Law In Springs

Statement: This Law states that the force required to stretch or compress a spring is linearly related to the change in its length.

           $F=-kx$

•  F =Force applied
• k = spring constant (a measure of the spring's stiffness, in Newtons per meter)
• x is the change in length of the spring from its equilibrium position (in meters)

The Negative Sign: This indicates that the force applied by the spring is in the reverse direction to the displacement. If you stretch the spring (positive x), the spring pulls back (negative F).  

Spring Constant (k): A larger spring constant means the spring is stiffer and requires more force to stretch or compress it.  

Elastic Limit: Hooke's Law only applies within the elastic limit of the spring. If you stretch or compress the spring too much, it will deform permanently, and Hooke's Law no longer holds.

Spring Constant(k)

• It is a measure of a spring's stiffness or resistance to deformation. It describes the amount of force required to stretch or compress a spring by a certain distance.
• The SI Unit of Spring constant(k) is $k=\frac{F}{x}\Rightarrow \frac{N}{m}$
•  A larger spring constant means the spring is stiffer, requiring more force to stretch or compress it by a given amount.
• A smaller spring constant means the spring is more flexible and requires less force to achieve the same amount of deformation.

9.0Solved Questions on Hooke’s Law

Q-1.Determine the effective spring constant for a system consisting of two springs, as shown in the figure.           

Solution:

When external force is applied, one spring gets extended and another one gets contracted by the same distance hence force due to two springs acting in the same direction.

$F=F_1+F_2\Rightarrow -kx=-k_{1}x-k_{2}x\Rightarrow k=k_1+k_2$

Q-2.The springs in question are identical. When a force of 4 kg (labeled as A) is applied, the spring stretches by 1 cm. If a force of 6 kg (labeled as B) is applied, how much will the spring elongate?

Solution:

$F=kx\Rightarrow mg=kx\Rightarrow m\propto kx$

$\frac{m_1}{m_2}=\frac{k_1}{k_2}\times \frac{x_1}{x_2}\Rightarrow \frac{4}{6}=\frac{k}{\frac{k}{2}}\times \frac{1}{x_1}\Rightarrow x_2=3Cm$

Q-3.Two wires made of the same material have radii in the ratio of 2:1. If the same force is applied to both, what is the ratio of the stress experienced by each wire?

Solution:

$\text{Stress}=\frac{\text{Force}}{\text{Area}}=\frac{F}{\pi r^2}$$\Rightarrow \frac{(Stress{)}_1}{(stress{)}_2}=\frac{F}{\pi r_1^2}\times \frac{\pi r_2^2}{F}$

$[\frac{r_2}{r_1}{]}^2=[\frac{1}{2}{]}^2=\frac{1}{4}$