To express [tex]$L$[/tex] in terms of [tex]$g$[/tex] and [tex]$f$[/tex], we start with the given equation for the period [tex]\( T \)[/tex] of a simple pendulum:
[tex]\[ T = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
We also know that the frequency [tex]\( f \)[/tex] is the reciprocal of the period [tex]\( T \)[/tex], so:
[tex]\[ f = \frac{1}{T} \][/tex]
Rewriting [tex]\( T \)[/tex] in terms of [tex]\( f \)[/tex]:
[tex]\[ T = \frac{1}{f} \][/tex]
Next, substitute [tex]\( T = \frac{1}{f} \)[/tex] into the original period equation:
[tex]\[ \frac{1}{f} = 2 \pi \sqrt{\frac{L}{g}} \][/tex]
To isolate [tex]\( L \)[/tex], we'll first get rid of the square root by squaring both sides of the equation:
[tex]\[ \left(\frac{1}{f}\right)^2 = (2 \pi)^2 \left(\frac{L}{g}\right) \][/tex]
Simplify the left side:
[tex]\[ \frac{1}{f^2} = 4 \pi^2 \left(\frac{L}{g}\right) \][/tex]
To isolate [tex]\( L \)[/tex], multiply both sides by [tex]\( g \)[/tex]:
[tex]\[ \frac{g}{f^2} = 4 \pi^2 L \][/tex]
Finally, divide both sides by [tex]\( 4 \pi^2 \)[/tex] to solve for [tex]\( L \)[/tex]:
[tex]\[ L = \frac{g}{4 \pi^2 f^2} \][/tex]
Thus, the correct answer is:
A. [tex]\( L = \frac{g}{4 \pi^2 f^2} \)[/tex]
