# Spring potential energy | definition, meaning and its derivation

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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.

# 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]

W_{c} = – \Delta U

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.
$$dw = \int -k.x.dx$$
$$\text{integrating both sides }$$
$$\int dw = -k \int_{0}^{x} x.dx$$

\implies W = – \frac {1}{2} kx^2

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

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

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.

W = – \frac{1}{2} kx^2
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

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

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

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