Answer-
[tex]\boxed{\boxed{L=\dfrac{g}{4\pi^2 f^2}}}[/tex]
Solution-
The equation for time period of a simple pendulum is given by,
[tex]T=2\pi \sqrt{\dfrac{L}{g}}[/tex]
Where,
T = Time period,
L = Length of the rod,
g = Acceleration due to gravity.
Frequency (f) of the pendulum is the reciprocal of its period, i.e
[tex]f=\dfrac{1}{T}\ \Rightarrow T=\dfrac{1}{f}[/tex]
Putting the values,
[tex]\Rightarrow \dfrac{1}{f}=2\pi \sqrt{\dfrac{L}{g}}[/tex]
[tex]\Rightarrow (\dfrac{1}{f})^2=(2\pi \sqrt{\dfrac{L}{g}})^2[/tex]
[tex]\Rightarrow \dfrac{1}{f^2}=4\pi^2 \dfrac{L}{g}[/tex]
[tex]\Rightarrow L=\dfrac{g}{4\pi^2 f^2}[/tex]
