The potential energy (U) stored in a spring when it is stretched or compressed is given by the formula:

$$U = \frac{1}{2} k x^2$$

where:

• $k$ is the spring constant,
• $x$ is the displacement from the spring's equilibrium position (i.e., how much the spring is stretched or compressed).

If the spring is initially stretched by $2$ cm (or $0.02$ meters, since we generally use SI units for these calculations), the potential energy stored can be represented as:

$$U = \frac{1}{2} k (0.02)^2$$

When the spring is stretched by $8$ cm (or $0.08$ meters), the new potential energy $U'$ becomes:

$$U' = \frac{1}{2} k (0.08)^2$$

To find the relation between $U'$ and $U$, we can divide the expression for $U'$ by that of $U$:

$$\frac{U'}{U} = \frac{\frac{1}{2} k (0.08)^2}{\frac{1}{2} k (0.02)^2}$$

Simplifying this quotient gives:

$$\frac{U'}{U} = \left(\frac{0.08}{0.02}\right)^2 = \left(4\right)^2 = 16$$

Thus, if the spring's displacement is increased from $2$ cm to $8$ cm, the potential energy stored in the spring increases by a factor of $16$, meaning:

$U' = 16U$