Hooke’s Law

Formula for Spring

Hooke’s law describes the forces in spring. A spring, in its natural form, is in an equilibrium position. When a force is applied to compress or extend it, the spring will restore its initial position. The force with which it restores is known as spring force and can be quantified using Hooke’s law, which states that the restoring force F is proportional to the spring’s displacement x.

F ∝ x

Or, F = -kx

SI unit of Hooke’s law: Newton or N (kg.m.s^-2)

Here, k is known as the force constant or spring constant. The negative sign is because the restoring force is in the opposite direction to the displacement. It is responsible for bringing back the spring to its equilibrium position. This equation applies to both compression and extension of spring. The spring force can be measured using a spring tester or a weighing scale. The magnitude of the force is given by,

F = kx

Spring Constant

The magnitude of the spring constant is given by,

k = F/x

Suppose F = 1 N and x = 1 m, then

k = 1 N/ 1 m = 1 N/m

Therefore, the spring constant is defined as the force required to displace the spring by one meter. It has a unit of Newton per meter (N/m) and a dimension given by MT^-2.

Elastic Energy

When spring is deformed, work is being done on it. This work is manifested in the form of elastic potential energy stored by the spring. By definition, work done is the change in potential energy. As the deformation is removed, energy is released as the potential energy is converted into kinetic energy. The work done W in compressing or extending the spring by a distance x is given by,

W = ΔPE = ½kx²

The ability of an object to resist deformation and restore to normal position is called elasticity. Hooke’s law is applied to a limited deformation known as the elastic limit. Suppose the object is compressed or extended beyond this limit. In that case, it will not restore to its original shape and will be permanently deformed.

Hooke’s law can be expressed graphically in which force is plotted against displacement. Since the two variables share a linear relationship, the graph will be a straight line.

Stress and strain are essential properties of ductile materials. Consider any object that is subjected to an applied force resulting in deformation. Stress is a quantity that describes the magnitude of the applied force. It is defined as the force applied per area of the object. Strain is used to describing the deformation. It is defined as the ratio of change in length (volume) to the object’s original length (volume). Hooke’s law gives a relationship between stress and strain.

Stress = Modulus of Elasticity x Strain

Stress-Strain Equation

An object can be deformed in different ways. The stress and strain can be of two types – 1. longitudinal or normal and 2. shear or torsion. Based on the types, Hooke’s law can be written as follows.

1. Normal or Longitudinal

σ = E x ε

Where,

σ is the normal or longitudinal stress

ε is the normal or longitudinal strain

E is Young’s modulus or modulus of elasticity

2. Shear or Torsion

τ = G x γ

Where,

σ is the shear or torsional stress

ε is the shear or torsional strain

G is the shear modulus

Hooke’s law can be applied to many devices like a mechanical clock, spring balance, watch, and manometer (pressure gauge). It has found its uses in various disciplines like acoustics, seismology, molecular mechanics, engineering, and construction.

P.1. How much force is needed to pull a spring with a spring constant of 15 N/m a distance of 15 cm?

Soln. Given,

k = 15 N/m

x = 15 cm = 0.15 m

Formula: F = kx

Therefore,

F = 15 N/m x 0.15 m = 2.25 N

P.2. A spring is pulled to 12 cm and held in place with a force of 550 N. What is the spring constant of the spring?

Soln. Given,

F = 550 N

x = 12 cm = 0.12 m

Formula: F = kx

Therefore,

k = F/x

or, k = 550 N/0.12 m = 4583 N/m

P.3. What is the force required to stretch a 15 cm-long spring, with a spring constant of 80 N/m, to a length of 17 cm?

Soln.: Given,

l₁ = 15 cm

l₂ = 17 cm

k = 80 N/m

Therefore,

x = l₂ – l₁ = 17 cm – 15 cm = 2 cm = 0.02 m

From Hooke’s law equation,

F = kx

F = 80 N/m x 0.02 m = 1.6 N

P.4. What is the spring constant of a spring that needs a force of 4 N to compress from 50 cm to 45 cm?

Soln.: Given,

F = 4 N

l₂ = 50 cm

l₁ = 45 cm

Therefore,

x = l₂ – l₁ = 45 cm – 50 cm = – 5 cm = – 0.05 m

or, |x| = 0.05 m

From Hooke’s law equation,

F = kx

or, k = F/x = 4N / 0.05 m = 80 N/m