Question
The formula to determine the period of one swing of a simple pendulum is T, equals, 2, pi, square root of, start fraction, L, divided by, g, end fraction, end square root, commaT=2π
g
L
, where LL is the length of the string and gg is the acceleration due to gravity. Solve the formula to solve for gg in terms of pi, commaπ, TT and L, .L.
Answer
Attempt 1 out of 2
g, equalsg=
g
L
, where LL is the length of the string and gg is the acceleration due to gravity. Solve the formula to solve for gg in terms of pi, commaπ, TT and L, .L.
Answer
Attempt 1 out of 2
g, equalsg=
Answers
Answered by
GPT-4o mini
To solve the formula for \( g \) in terms of \( T \) and \( L \), we start with the original formula:
\[
T = 2\pi \sqrt{\frac{L}{g}}
\]
We want to isolate \( g \). Here are the steps to do so:
1. Square both sides to eliminate the square root:
\[
T^2 = (2\pi)^2 \frac{L}{g}
\]
2. Simplify the right side:
\[
T^2 = 4\pi^2 \frac{L}{g}
\]
3. Rearrange to solve for \( g \):
\[
g = 4\pi^2 \frac{L}{T^2}
\]
Thus, the final expression for \( g \) is:
\[
g = \frac{4\pi^2 L}{T^2}
\]
\[
T = 2\pi \sqrt{\frac{L}{g}}
\]
We want to isolate \( g \). Here are the steps to do so:
1. Square both sides to eliminate the square root:
\[
T^2 = (2\pi)^2 \frac{L}{g}
\]
2. Simplify the right side:
\[
T^2 = 4\pi^2 \frac{L}{g}
\]
3. Rearrange to solve for \( g \):
\[
g = 4\pi^2 \frac{L}{T^2}
\]
Thus, the final expression for \( g \) is:
\[
g = \frac{4\pi^2 L}{T^2}
\]
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