Question

Use the table to answer the question.
Oscillation 1 2 3 4 5
Distance 12
A pendulum swinging back and forth loses momentum and distance with each oscillation. The
total distance the pendulum swings on the first oscillation is 12 inches, and the distance of each
consecutive oscillation is
2
3
of the previous distance. What distance will the pendulum travel on the
fifth oscillation, measured to the nearest whole inch?

In the fifth oscillation, the pendulum will travel approximately inches.

Answer 1

To find the distance the pendulum travels in the fifth oscillation, we can use the information given about the distance for each consecutive oscillation being \( \frac{2}{3} \) of the previous distance.

1. For the first oscillation:
\[
d_1 = 12 \text{ inches}
\]

2. For the second oscillation:
\[
d_2 = d_1 \times \frac{2}{3} = 12 \times \frac{2}{3} = 8 \text{ inches}
\]

3. For the third oscillation:
\[
d_3 = d_2 \times \frac{2}{3} = 8 \times \frac{2}{3} = \frac{16}{3} \approx 5.33 \text{ inches}
\]

4. For the fourth oscillation:
\[
d_4 = d_3 \times \frac{2}{3} = \frac{16}{3} \times \frac{2}{3} = \frac{32}{9} \approx 3.56 \text{ inches}
\]

5. For the fifth oscillation:
\[
d_5 = d_4 \times \frac{2}{3} = \frac{32}{9} \times \frac{2}{3} = \frac{64}{27} \approx 2.37 \text{ inches}
\]

Now rounding \( d_5 \) to the nearest whole inch:
\[
d_5 \approx 2.37 \approx 2 \text{ inches}
\]

Thus, in the fifth oscillation, the pendulum will travel approximately **2 inches**.