The period (in seconds) of a simple pendulum is related to its length (in meters) and the acceleration due to gravity (in meters per second squared) by the formula What is in terms of and ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 137 | Free SAT Prep | Aniko

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

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The period $T$ (in seconds) of a simple pendulum is related to its length $L$ (in meters) and the acceleration due to gravity $g$ (in meters per second squared) by the formula

T = 2\pi\sqrt{\frac{L}{g}}

What is $g$ in terms of $T$ and $L$ ?

For pendulum or physics-style formulas on the SAT where you must solve for a different variable, treat it like a regular algebra problem: first isolate any radical or complicated part (divide away constants), then eliminate the radical (by squaring) or fraction step by step, and finally use cross-multiplication to solve for the target variable. At each stage, keep track of which quantities end up in the numerator or denominator—this helps you quickly rule out options that reverse the proportionality (for example, putting $T^2$ in the wrong place).

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Start by isolating the radical

Notice that $T$ is multiplied by $2\pi$ and then by a square root. What can you do first to get the square root by itself?

Clear the square root

Once the square root is isolated, think about what operation will remove a square root from an equation.

Solve the resulting fraction equation

After squaring both sides, you will get a fraction involving $L$ and $g$ . How can you rearrange that fraction to get $g$ by itself? Consider cross-multiplying or inverting carefully.

Check where T and L end up

In your final expression, ask: Is $g$ directly proportional to $L$ or to $T^2$ ? Should $T^2$ be in the numerator or denominator based on the original equation?

Desmos Guide

Pick convenient test values for L and g

In Desmos, type something like L = 3 and g = 5 on separate lines to define specific values for $L$ and $g$ .

Compute T from the original formula

On a new line, enter T = 2*pi*sqrt(L/g) so Desmos calculates the corresponding value of $T$ using your chosen $L$ and $g$ .

Test each answer choice numerically

For each choice, type its expression for $g$ as a new line in Desmos, using the $T$ and $L$ already defined (for example, gA = T^2/(4*pi^2*L), gB = 4*pi^2*T^2/L, etc.). Compare each computed $g$ value with your original g from step 1 and see which expression reproduces that original number.

Confirm the correct symbolic form

The only answer choice whose computed $g$ matches the original $g$ you set will be the correct algebraic rearrangement of the formula.

Step-by-step Explanation

Write the given formula and identify the goal

We are given

T = 2\pi\sqrt{\frac{L}{g}}

and we want to solve for $g$ in terms of $T$ and $L$ . That means we need to rearrange the equation so that $g$ is alone on one side.

Isolate the square root

First, divide both sides by $2\pi$ to get rid of the coefficient in front of the square root:

\frac{T}{2\pi} = \sqrt{\frac{L}{g}}.

Now the square root is alone on the right side.

Square both sides to remove the square root

To eliminate the square root, square both sides of the equation:

\left(\frac{T}{2\pi}\right)^2 = \left(\sqrt{\frac{L}{g}}\right)^2.

This simplifies to

\frac{T^2}{4\pi^2} = \frac{L}{g}.

Now we have a fraction equation without radicals.

Solve the fraction equation for g

From

\frac{T^2}{4\pi^2} = \frac{L}{g},

we want $g$ alone. One clean way is to cross-multiply:

T^2 \cdot g = 4\pi^2 \cdot L.

Now divide both sides by $T^2$ to isolate $g$ :

g = \frac{4\pi^2 L}{T^2}.

So the correct expression for $g$ in terms of $T$ and $L$ is $g = \dfrac{4\pi^2 L}{T^2}$ , which corresponds to choice C.

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation