Answer :
We are given the formula for the period of a pendulum:
[tex]$$
T = 2\pi \sqrt{\frac{L}{32}},
$$[/tex]
with the value [tex]$T = 1.57$[/tex] seconds and [tex]$\pi = 3.14$[/tex].
Step 1. Isolate the square root term by dividing both sides by [tex]$2\pi$[/tex]:
[tex]$$
\sqrt{\frac{L}{32}} = \frac{T}{2\pi} = \frac{1.57}{2 \times 3.14}.
$$[/tex]
Step 2. Calculate the right-hand side:
[tex]$$
\frac{1.57}{6.28} = 0.25.
$$[/tex]
So we have:
[tex]$$
\sqrt{\frac{L}{32}} = 0.25.
$$[/tex]
Step 3. Square both sides of the equation to remove the square root:
[tex]$$
\frac{L}{32} = (0.25)^2 = 0.0625.
$$[/tex]
Step 4. Multiply both sides by 32 to solve for [tex]$L$[/tex]:
[tex]$$
L = 32 \times 0.0625 = 2.
$$[/tex]
Thus, the length of the pendulum is [tex]$2$[/tex] feet.
Final Answer: The length is [tex]$\boxed{2 \text{ feet}}$[/tex].
[tex]$$
T = 2\pi \sqrt{\frac{L}{32}},
$$[/tex]
with the value [tex]$T = 1.57$[/tex] seconds and [tex]$\pi = 3.14$[/tex].
Step 1. Isolate the square root term by dividing both sides by [tex]$2\pi$[/tex]:
[tex]$$
\sqrt{\frac{L}{32}} = \frac{T}{2\pi} = \frac{1.57}{2 \times 3.14}.
$$[/tex]
Step 2. Calculate the right-hand side:
[tex]$$
\frac{1.57}{6.28} = 0.25.
$$[/tex]
So we have:
[tex]$$
\sqrt{\frac{L}{32}} = 0.25.
$$[/tex]
Step 3. Square both sides of the equation to remove the square root:
[tex]$$
\frac{L}{32} = (0.25)^2 = 0.0625.
$$[/tex]
Step 4. Multiply both sides by 32 to solve for [tex]$L$[/tex]:
[tex]$$
L = 32 \times 0.0625 = 2.
$$[/tex]
Thus, the length of the pendulum is [tex]$2$[/tex] feet.
Final Answer: The length is [tex]$\boxed{2 \text{ feet}}$[/tex].
