Answer :
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]$\pi \approx 3.14$[/tex], and [tex]$L$[/tex] is the length in feet.
1. Substitute the given period [tex]$T = 1.57$[/tex] seconds into the formula:
[tex]$$
1.57 = 2 \pi \sqrt{\frac{L}{32}}.
$$[/tex]
2. Divide both sides by [tex]$2\pi$[/tex] to isolate the square root:
[tex]$$
\sqrt{\frac{L}{32}} = \frac{1.57}{2\pi}.
$$[/tex]
Given that [tex]$\pi = 3.14$[/tex],
[tex]$$
\sqrt{\frac{L}{32}} = \frac{1.57}{2 \times 3.14} = \frac{1.57}{6.28} = 0.25.
$$[/tex]
3. Square both sides of the equation to remove the square root:
[tex]$$
\left(\sqrt{\frac{L}{32}}\right)^2 = (0.25)^2,
$$[/tex]
which simplifies to
[tex]$$
\frac{L}{32} = 0.0625.
$$[/tex]
4. Multiply both sides of the equation by [tex]$32$[/tex] 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.
[tex]$$
T = 2 \pi \sqrt{\frac{L}{32}},
$$[/tex]
where [tex]$T$[/tex] is the period in seconds, [tex]$\pi \approx 3.14$[/tex], and [tex]$L$[/tex] is the length in feet.
1. Substitute the given period [tex]$T = 1.57$[/tex] seconds into the formula:
[tex]$$
1.57 = 2 \pi \sqrt{\frac{L}{32}}.
$$[/tex]
2. Divide both sides by [tex]$2\pi$[/tex] to isolate the square root:
[tex]$$
\sqrt{\frac{L}{32}} = \frac{1.57}{2\pi}.
$$[/tex]
Given that [tex]$\pi = 3.14$[/tex],
[tex]$$
\sqrt{\frac{L}{32}} = \frac{1.57}{2 \times 3.14} = \frac{1.57}{6.28} = 0.25.
$$[/tex]
3. Square both sides of the equation to remove the square root:
[tex]$$
\left(\sqrt{\frac{L}{32}}\right)^2 = (0.25)^2,
$$[/tex]
which simplifies to
[tex]$$
\frac{L}{32} = 0.0625.
$$[/tex]
4. Multiply both sides of the equation by [tex]$32$[/tex] 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.
