College

The period [tex]$T$[/tex] (in seconds) of a pendulum is given by [tex]$T = 2 \pi \sqrt{\frac{L}{32}}$[/tex], where [tex]$L$[/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. 8 feet
D. 2 feet

Answer :

Let's solve the problem step by step:

We are given the formula for the period [tex]\( T \)[/tex] of a pendulum:

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

where:
- [tex]\( T \)[/tex] is the period in seconds,
- [tex]\( L \)[/tex] is the length of the pendulum in feet.

We know:
- [tex]\( T = 1.57 \)[/tex] seconds,
- [tex]\( \pi = 3.14 \)[/tex].

We are asked to find [tex]\( L \)[/tex].

1. Rearrange the formula to solve for [tex]\( L \)[/tex]:

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

Divide both sides by [tex]\( 2 \pi \)[/tex]:

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

2. Square both sides to eliminate the square root:

[tex]\[
\left(\frac{T}{2 \pi}\right)^2 = \frac{L}{32}
\][/tex]

3. Multiply both sides by 32 to solve for [tex]\( L \)[/tex]:

[tex]\[
L = 32 \times \left(\frac{T}{2 \pi}\right)^2
\][/tex]

4. Substitute in the known values of [tex]\( T \)[/tex] and [tex]\( \pi \)[/tex]:

[tex]\[
L = 32 \times \left(\frac{1.57}{2 \times 3.14}\right)^2
\][/tex]

5. Calculate inside the parentheses first:

[tex]\[
\frac{1.57}{2 \times 3.14} = \frac{1.57}{6.28} \approx 0.25
\][/tex]

6. Square the result:

[tex]\[
0.25^2 = 0.0625
\][/tex]

7. Multiply by 32:

[tex]\[
L = 32 \times 0.0625 = 2
\][/tex]

Therefore, the length of the pendulum is [tex]\( 2 \)[/tex] feet.