Question 137·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

No Desmos needed. Rearranging by hand gives a formula that works for every pendulum. The question asks for $g$ by itself, so treat $T$ and $L$ as fixed values. First, divide both sides of the given equation by $2\pi$:

A square undoes a square root. Square both whole sides, including the $2\pi$ beneath $T$:

The fraction on the right means $L$ divided by $g$. Multiply both sides by $g$ to move it out of the denominator:

The factor multiplying $g$ is $T^2/(4\pi^2)$. Since a period $T$ is positive, multiply both sides by its reciprocal, $4\pi^2/T^2$:

So the acceleration due to gravity is $4\pi^2L/T^2$ in terms of $T$ and $L$. Choice C.

Because the choices are formulas, one test case can rule out a wrong formula, though a match alone doesn’t prove one works for every case. Let $L=1$ and $T=2$. Type those values, then type $T\sim2\pi\sqrt{L/g_1}\{g_1>0\}$. The $\sim$ asks Desmos to fit the equation, and the subscript makes $g_1$ the value to find. Desmos shows $g_1\approx9.8696$ under PARAMETERS.

Type each choice’s expression using the same $T$ and $L$. Desmos shows approximately $0.1013$, $157.9137$, $9.8696$, and $0.0063$, in that order. Only $4\pi^2L/T^2$ matches the $g_1$ found from the given equation. Since each other choice fails this test case, the matching formula is the answer. Choice C.