The period (in seconds) of a simple pendulum of length (in meters) is given by where is the acceleration due to gravity, in meters per second squared. What is in terms of and ?

Solution available with step-by-step explanation

SAT Nonlinear Equations & Systems Question 110 | Free SAT Prep | Aniko

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Question 110·Medium·Nonlinear Equations in One Variable; Systems in Two Variables

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The period $T$ (in seconds) of a simple pendulum of length $L$ (in meters) is given by

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

where $g$ is the acceleration due to gravity, in meters per second squared.

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

When a formula question asks for one variable in terms of others, treat it like a normal algebra equation: isolate the part with the variable (here the square root containing L), use inverse operations step by step (divide, then square to remove the root), and carefully track which quantities are in the numerator or denominator. After simplifying, solve the resulting fraction equation for the desired variable and quickly compare with the answer choices, paying close attention to where each symbol (like g) ends up.

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Get the square root alone

Start by dividing both sides of $T = 2\pi \sqrt{L/g}$ by $2\pi$ so that the square root is by itself on one side.

Eliminate the square root

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

Solve the resulting fraction equation

After you square both sides, you will get a fraction involving $L$ and $g$ . Rearrange that equation so that $L$ is alone on one side.

Desmos Guide

Enter the original formula with sliders

Type T = 2*pi*sqrt(L/g) into Desmos. When prompted, create sliders for T, L, and g, and set T and g to some positive values (for example, T = 3 and g = 9.8).

Test each answer choice as a formula for L

For each answer choice, type a separate expression defining L in terms of T and g, such as L1 = g*T^2/(4*pi^2), L2 = T^2/(4*pi^2*g), and so on, using the exact forms from the options.

Check which formula makes the equation true

For each defined Li, type T - 2*pi*sqrt(Li/g) and look at the numeric result. The correct formula for L is the one that makes this expression equal to 0 (or extremely close to 0) for your chosen values of T and g.

Step-by-step Explanation

Isolate the square root

We are given

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

We want to solve for $L$ . First, divide both sides by $2\pi$ to isolate the square root:

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

Remove the square root by squaring

Square both sides of the equation to get rid of the square root:

\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}.

Solve for L

Now multiply both sides by $g$ to solve for $L$ :

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

So the expression for $L$ in terms of $T$ and $g$ is $\dfrac{g T^2}{4\pi^2}$ .

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Step-by-step Explanation

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Step-by-step Explanation