To determine the correct length of the pendulum, we will use the given formula and the known period of the pendulum’s swing.
The formula relating the period [tex]\( T \)[/tex] of a pendulum to its length [tex]\( L \)[/tex] is:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
where [tex]\( T \)[/tex] is the period, [tex]\( L \)[/tex] is the length of the pendulum, [tex]\( \pi \)[/tex] is approximately 3.14159, and [tex]\( g \)[/tex] is the acceleration due to gravity, which is 980 cm/s² in this context.
We are given:
- The period [tex]\( T \)[/tex] is 0.7 seconds.
- [tex]\( g \)[/tex] is 980 cm/s².
To find [tex]\( L \)[/tex], we rearrange the formula to solve for [tex]\( L \)[/tex]:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{980}} \][/tex]
First, isolate the square root term by dividing both sides by [tex]\( 2 \pi \)[/tex]:
[tex]\[ \frac{T}{2 \pi} = \sqrt{\frac{L}{980}} \][/tex]
Next, square both sides to remove the square root:
[tex]\[ \left( \frac{T}{2 \pi} \right)^2 = \frac{L}{980} \][/tex]
Then, solve for [tex]\( L \)[/tex] by multiplying both sides by 980:
[tex]\[ L = 980 \left( \frac{T}{2 \pi} \right)^2 \][/tex]
Substitute [tex]\( T = 0.7 \)[/tex] seconds into the equation:
[tex]\[ L = 980 \left( \frac{0.7}{2 \pi} \right)^2 \][/tex]
Next, calculate [tex]\( \frac{0.7}{2 \pi} \)[/tex]:
[tex]\[ \frac{0.7}{2 \pi} = \frac{0.7}{6.28318} \approx 0.11148 \][/tex]
Then, square this value:
[tex]\[ (0.11148)^2 \approx 0.01243 \][/tex]
Finally, multiply by 980 to get [tex]\( L \)[/tex]:
[tex]\[ L = 980 \times 0.01243 \approx 12.163 \][/tex]
Rounding 12.163 to the nearest ones place, we get:
[tex]\[ L \approx 12 \][/tex]
Therefore, the length of the new pendulum should be approximately 12 centimeters. The correct answer is:
D. 12 centimeters
