The angular frequency of the damped oscillator is given by, $$\omega = \sqrt{\left(\frac{k}{m} - \frac{r^2}{4m^2}\right)}$$ where k is the spring constant, m is the mass of the oscillator and r is the damping constant. If the ratio $$\frac{r^2}{mk}$$ is 8%, the change in time period compared to the undamped oscillator is approximately as follows:

The angular frequency of the undamped oscillator is $$\omega_0 = \sqrt{\frac{k}{m}}$$.

$$\omega = \sqrt{\frac{k}{m} \left( 1 - \frac{r^2}{4mk} \right)} = \omega_0 \left( 1 - \frac{r^2}{4mk} \right)^{1/2}$$

$$T = \frac{2\pi}{\omega_0 \left( 1 - \frac{r^2}{4mk} \right)^{1/2}} = T_0 \left( 1 - \frac{r^2}{4mk} \right)^{-1/2}$$, where $$T_0$$ is the time period of the undamped oscillator.

Given that the ratio $$\frac{r^2}{mk} = 8\% = 0.08$$, the term $$\frac{r^2}{4mk} = \frac{0.08}{4} = 0.02$$. Since $$0.02 \ll 1$$, we can use the binomial approximation $$(1 - x)^n \approx 1 + nx$$: $$T \approx T_0 \left[ 1 + \left( -\frac{1}{2} \right) \left( -\frac{r^2}{4mk} \right) \right]$$

$$T \approx T_0 \left( 1 + \frac{r^2}{8mk} \right)$$

$$\frac{\Delta T}{T_0} = \frac{T - T_0}{T_0} = \frac{r^2}{8mk}$$

$$\text{Percentage change} = \left( \frac{r^2}{8mk} \right) \times 100 = \frac{1}{8} \left( \frac{r^2}{mk} \times 100 \right)$$

$$\frac{r^2}{mk} \times 100 = 8\%$$

$$\text{Percentage change} = \frac{1}{8} \times 8\% = 1\%$$