We that know how **potential energy** is defined for any **conservative forces**, here we will derive the Potential energy of **spring**. All the results are derived assuming **ideal springs**.

Inside Story

# The potential energy of Spring

### Meaning of potential energy of spring

We can see that if we compress or elongate a **spring**, we feel opposing force by spring, Whenever we disturb spring from the natural length it tries to retain its natural length, therefore there is some energy stored in a compressed or stretched spring due to which it tries to retain natural length.

Actually, this energy stored in any stretched/compressed spring or in stretching of any rubber band, etc, is elastic potential energy as the bodies are resisting change in its shape, size, length, etc.

### Derivation of the potential energy of spring

#### When compressing a spring from the natural length

**Spring force** is conservative therefore by **conservative field theory** we know that

[Latexpage]

\begin{equation}

W_{c} = – \Delta U

\end{equation}

Suppose we have a spring of natural length “l”. We compress it by x length from the natural length, now in this compressed state spring is capable of doing **work** hence having energy.

Now **work done** by spring in compressing

$$

dw = \int \vec{F}.\vec{dx}

$$

$$

dw = \int |\vec{F}| |\vec{dx}| \cos{\theta} $$

$$ \theta = 180 $$

Because force and displacement are opposite.

$$

dw = \int -k.x.dx

$$

$$ \text{integrating both sides }

$$

$$

\int dw = -k \int_{0}^{x} x.dx

$$

\begin{equation}

\implies W = – \frac {1}{2} kx^2

\end{equation}

From equation (1) and (2)

$$

– \frac {1}{2} kx^2 = – \Delta U $$

$$

\implies

\frac {1}{2} kx^2 = U_{final} – U_{initial}

$$

$$

\implies

\frac {1}{2} kx^2 = U_{x = x} – U_{x = 0}

$$

If we take x = 0 i.e. natural length and take it as a reference point and assume the **potential energy** at natural length to be zero, then

\begin{equation}

\frac {1}{2} kx^2 = U_{x} \end{equation}

compress a spring by an amount “x” w.r.t natural length the potential energy stored in spring at that state is given by equation (3).

#### When stretching a spring from natural length.

We know that whether we stretch or compress a spring by “x” amount from the natural length the **work done by spring in both cases is the same.**

Therefore if we strech spring by “x” amount.

\begin{equation}

W = – \frac{1}{2} kx^2 \end{equation}

From equation (1) and (4) .

$$

– \frac {1}{2} kx^2 = – \Delta U $$

$$

\implies

\frac {1}{2} kx^2 = U_{final} – U_{initial}

$$

$$

\implies

\frac {1}{2} kx^2 = U_{x = x} – U_{x = 0}

$$

If we take x = 0 i.e. natural length and take it as a reference point and assume the **potential energy** at natural length to be zero, then

\begin{equation}

\frac {1}{2} kx^2 = U_{x} \end{equation}

Stretch a spring by an amount “x” w.r.t natural length the potential energy stored in spring at that state is given by equation (5).

# Result

From equations (3) and (5), we can see that whether we compress or stretch a spring by an amount “x” the amount of potential energy of spring in that state is the same in both springs.

Given by

\begin{equation}

U_{x} = \frac{1}{2}kx^2

\end{equation}