1. **Problem statement:** A pendulum has a period of 3.0 s as measured by an observer on the Earth moving with a speed of 0.80 c relative to the pendulum's frame. Find the period of the pendulum measured by the observer. 2. **Relevant formula:** The time dilation formula from special relativity is used here: $$T = T_0 \gamma = \frac{T_0}{\sqrt{1 - \frac{v^2}{c^2}}}$$ where $T_0$ is the proper period (period in the pendulum's rest frame), $T$ is the dilated period measured by the moving observer, $v$ is the relative velocity, and $c$ is the speed of light. 3. **Given:** - $T = 3.0$ s (period measured by moving observer) - $v = 0.80 c$ 4. **Find:** $T_0$ (period in pendulum's rest frame) 5. **Rearrange the formula to find $T_0$:** $$T_0 = T \sqrt{1 - \frac{v^2}{c^2}}$$ 6. **Calculate:** $$\sqrt{1 - (0.80)^2} = \sqrt{1 - 0.64} = \sqrt{0.36} = 0.6$$ $$T_0 = 3.0 \times 0.6 = 1.8 \text{ s}$$ 7. **Interpretation:** The proper period of the pendulum (its period in its own rest frame) is 1.8 s. **Answer:** C. 1.8 s