To calculate the total distance the pendulum travels by the 5th oscillation, we start with the first oscillation. The pendulum swings a distance of 12 inches on the first oscillation. Each subsequent oscillation travels \( \frac{2}{3} \) of the distance of the previous oscillation.
We can represent the distances for each oscillation as follows:
- 1st oscillation: \( d_1 = 12 \) inches
- 2nd oscillation: \( d_2 = \frac{2}{3} \cdot d_1 = \frac{2}{3} \cdot 12 = 8 \) inches
- 3rd oscillation: \( d_3 = \frac{2}{3} \cdot d_2 = \frac{2}{3} \cdot 8 = \frac{16}{3} \approx 5.33 \) inches
- 4th oscillation: \( d_4 = \frac{2}{3} \cdot d_3 = \frac{2}{3} \cdot \frac{16}{3} = \frac{32}{9} \approx 3.56 \) inches
- 5th oscillation: \( d_5 = \frac{2}{3} \cdot d_4 = \frac{2}{3} \cdot \frac{32}{9} = \frac{64}{27} \approx 2.37 \) inches
Now, we calculate the distances numerically:
- 1st oscillation: 12 inches
- 2nd oscillation: 8 inches
- 3rd oscillation: \( \frac{16}{3} \) inches ≈ 5.33 inches
- 4th oscillation: \( \frac{32}{9} \) inches ≈ 3.56 inches
- 5th oscillation: \( \frac{64}{27} \) inches ≈ 2.37 inches
Now, rounding the 5th oscillation distance to the nearest whole inch:
- The distance for the 5th oscillation is approximately 2.37 inches, which rounds to 2 inches.
Thus, the total distance the pendulum will travel on the 5th oscillation, rounded to the nearest whole inch, is 2 inches.