Hooke's Law

In a nutshell

Hooke's law is a proportionality principle in physics that states that up until the elastic limit of a material, the displacement of its extension or compression is directly proportional to the force applied on the material.

Equations

DESCRIPTION	EQUATION
Hooke's Law	$F=kx$
Springs in series	$\dfrac {1}{k_ {total}} = \dfrac {1}{k_1} + \dfrac {1}{k_2}$
Springs in parallel	$k_ {total} = k_1 + k_2$
Elastic potential energy	$E = \dfrac {1}{2} Fx$ $E = \dfrac {1}{2} kx^2$

Variable definitions

QUANTITY NAME	SYMBOL	DERIVED UNIT	ALTERNATE UNIT	SI BASE UNIT
$force$	$F$	$N$		$kg \, m \, s^{-2}$
$displacement$	$x$	$m$		$m$
$elastic\ potential\ energy$	$E$	$J$	$Ws$	$m^{2} \, kg \, s^{-2}$
$force \space (spring) \space constant$	$k$	$N \, m^{-1}$

Hooke's Law

Hooke's law is an important proportionality relation in physics that states that the displacement of a material's compression is directly proportional to the force applied to the material, until the material reaches the limit of proportionality. These two values are linked by the force constant ( $k$ ), which can also be called the spring constant when referring to springs.

The force constant is used to evaluate how stiff a material is. The larger the value of the force constant, the more difficult it is to deform.

From a point called the limit of proportionality, Hooke's Law will not apply. This is because the displacement of the material will no longer be proportional to the force exerted..

Hooke's law can be demonstrated through a force-extension graph.

Physics; Materials; KS5 Year 12; Hooke's law

A.	$force \ (N)$
B.	$extension \ (m)$
1.	Limit of proportionality

Example

A mass weighing $20\ kg$ is attached to one end of spring which has an original length of $1.0\ m$ . The new length of the spring is $1.2\ m$ . What is the spring constant of the spring used?

Firstly, write down the known values:

$m=20\ kg$ 

$x=1.2\ m - 1\ m = 0.2 \, m$ 

Next, calculate $F$ by using $F = mg$ :

$F = 20 × 9.81$

$F = 196.2 \ N$ 

Rearrange the equation for Hooke's law to make the spring constant the subject:

$k = \dfrac {F}{x}$

Substitute in values:

$k = \dfrac {196.2}{0.2}$ 

$k = 981\ Nm^{-1}$

The spring constant of the spring used is $\underline{980\ Nm^{-1}}$ (2.s.f).

Springs in series

Springs in series, means springs one after the other, just like components in a series circuit.

Physics; Materials; KS5 Year 12; Hooke's law

When a weight is added onto springs in series, the total displacement of the whole system is:

x_{total} = x_1+x_2+x_3

Using Hooke's law,

x

can be replaced with

\dfrac{F}{k}

:

\dfrac{F_{total}}{k_{total}} = \dfrac{F_1}{k_{1}}+ \dfrac{F_2}{k_{2}}+\dfrac{F_3}{k_{3}}

The force on each of the springs is the same so:

F_{total} = F_1, F_2, F_3

Therefore, the total spring constant of a system of springs in series is:

\dfrac {1}{k_ {total}} = \dfrac {1}{k_1} + \dfrac {1}{k_2}

Example

Two different springs, one with a spring constant of $6.8 \ Nm^{-1}$  and another with a spring constant of $2.6 \ Nm^{-1}$ are arranged in a series placement. What is the total spring constant of this system?

Firstly, write down the known values:

$k_1 = 6.8\ Nm^{-1}$

$k_2 = 2.6\ Nm^{-1}$

Write down the equation for calculating spring constant in series:

$\dfrac {1}{k_ {total}} = \dfrac {1}{k_1} + \dfrac {1}{k_2}$

Substitute the known values into the equation:

$\dfrac {1}{k_ {total}} = \dfrac {1}{6.8} + \dfrac {1}{2.6}$

$\dfrac {1}{k_ {total}} = 0.5317 ...$

Find the reciprocal to get $k_{total}$ :

$k_{total} = \dfrac {1}{0.5317...} = 1.88 \, N \, m^{-1}$ 

The total spring constant of the system is $\underline{1.9\ Nm^{-1}}$ (2.s.f).

Springs in parallel

For springs in parallel, the force is shared between the springs.

Physics; Materials; KS5 Year 12; Hooke's law

This means that:

$F_{total} = F_1 +F_2+F_3$

Substituting $kx$ in for $F$ :

$k_{total}x_{total}= k_{1}x_{1}+k_{2}x_{2}+k_{3}x_{3}$

The springs will all be extended the same amount so:

$x_{total} = x_1, x_2, x_3$

Therefore, the total spring constant of a system of springs in parallel is:

$k_ {total} = k_1 + k_2$

Example

Two different springs, one with a spring constant of $7.9 \ Nm^{-1}$ and another with a spring constant of $5.4 \ Nm^{-1}$ are arranged in a parallel placement. What is the total spring constant of the system?

Firstly, write down the known values:

$k_1 = 7.9\ Nm^{-1}$

$k_2 = 5.4\ Nm^{-1}$

Substitute the known values into the equation:

$k_{total} = 7.9 + 5.4$

 $k_{total} = 13.3\ Nm^{-1}$

The total spring constant of the system is $\underline{13\ Nm^{-1}}$ (2.s.f).

Elastic potential energy

When a material is extended or compressed, work done on a material is transferred to elastic potential energy. This energy is represented as the area under the graph in a force-extension graph, and therefore can be written as follows:

$E = \dfrac {1}{2} Fx$

This equation can be written in another form as we know that $F = kx$

$E = \dfrac {1}{2} Fx = \dfrac {1}{2} (kx) × x$

$E = \dfrac {1}{2} kx^2$

Example

The elastic potential energy of a spring that is extended by $0.60\ m$ is $48\ J$ . Calculate the spring constant of the spring.

Firstly, write down the known values:

$x = 0.6\ m$ 

$E = 48\ J$ 

Next, write down the equation for elastic potential energy:

$E = \dfrac {1}{2} kx^2$ 

Rearrange the equation to make $k$ the subject:

$k = \dfrac {2E}{x^2}$ 

Substitute the known values into the rearranged equation:

$k = \dfrac {2 ×48}{0.6^2}$

$k = 266.67 \ Nm^{-1}$

The spring constant is $\underline{270\ Nm^{-1}}$ (2.s.f).