A pendulum is L meters long. The time, t, in seconds that it takes to swing back and forth once is given by t = 2/L. Suppose the pendulum takes 7.26

seconds to swing back and forth once. What is its length?
Carry your intermediate computations to at least four decimal places, and round your answer to the nearest tenth.

1 answer

The formula for the time \( t \) of a pendulum swinging back and forth is given by:

\[ t = 2 \sqrt{\frac{L}{g}} \]

where \( g \) is the acceleration due to gravity, approximately \( 9.81 , \text{m/s}^2 \). However, based on your wording of the problem as \( t = 2/L \), I will assume you meant \( t = 2\pi \sqrt{\frac{L}{g}} \).

Given that \( t = 7.26 \) seconds, we can rearrange the formula to find the length \( L \):

\[ 7.26 = 2 \pi \sqrt{\frac{L}{g}} \]

First, we can solve for \( \sqrt{\frac{L}{g}} \):

\[ \sqrt{\frac{L}{g}} = \frac{7.26}{2\pi} \]

Calculating \( 2\pi \):

\[ 2\pi \approx 6.2832 \]

Now, we substitute this into our equation:

\[ \sqrt{\frac{L}{g}} = \frac{7.26}{6.2832} \approx 1.1560 \]

Now, squaring both sides:

\[ \frac{L}{g} = (1.1560)^2 \approx 1.3345 \]

Now, multiplying by \( g \) to solve for \( L \):

\[ L = 1.3345 \cdot 9.81 \approx 13.1074 \]

Finally, rounding \( L \) to the nearest tenth:

\[ L \approx 13.1 , \text{meters} \]

Thus, the length of the pendulum is approximately 13.1 meters.