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The period [tex]$T$[/tex] (in seconds) of a pendulum is given by [tex]$T = 2 \pi \sqrt{\frac{L}{32}}$[/tex], where [tex][tex]$L$[/tex][/tex] stands for the length (in feet) of the pendulum. If [tex]$\pi = 3.14$[/tex], and the period is 1.57 seconds, what is the length?

A. 16 feet
B. 20 feet
C. 2 feet
D. 8 feet

We are given the formula for the period of a pendulum:

[tex]$$
T = 2\pi \sqrt{\frac{L}{32}},
$$[/tex]

where:
- [tex]$T$[/tex] is the period (in seconds),
- [tex]$\pi = 3.14$[/tex] (approximation), and
- [tex]$L$[/tex] is the length of the pendulum (in feet).

Given that [tex]$T = 1.57$[/tex] seconds, we need to solve for [tex]$L$[/tex]. Follow these steps:

1. Substitute the given value into the equation:

[tex]$$
1.57 = 2 \cdot 3.14 \cdot \sqrt{\frac{L}{32}}.
$$[/tex]

2. Simplify the constant term:

Compute [tex]$2 \cdot 3.14$[/tex]:

[tex]$$
2 \cdot 3.14 = 6.28.
$$[/tex]

Now the equation becomes:

[tex]$$
1.57 = 6.28 \sqrt{\frac{L}{32}}.
$$[/tex]

3. Isolate the square root term:

Divide both sides by [tex]$6.28$[/tex]:

[tex]$$
\sqrt{\frac{L}{32}} = \frac{1.57}{6.28}.
$$[/tex]

Notice that:

[tex]$$
\frac{1.57}{6.28} = 0.25.
$$[/tex]

4. Eliminate the square root by squaring both sides:

[tex]$$
\left(\sqrt{\frac{L}{32}}\right)^2 = (0.25)^2,
$$[/tex]

which simplifies to:

[tex]$$
\frac{L}{32} = 0.0625.
$$[/tex]

5. Solve for [tex]$L$[/tex]:

Multiply both sides by [tex]$32$[/tex]:

[tex]$$
L = 32 \times 0.0625.
$$[/tex]

Calculating this:

[tex]$$
L = 2.
$$[/tex]

Thus, the length of the pendulum is [tex]$\boxed{2\ \text{feet}}$[/tex].