Question

22. What is the length of a pendulum that has a period of 0.500 s?

Answer 1

6.21 cm

• use the pendulum formula : $\sf \bold{\mathrm{T}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{g}}}}$

where

• T is time or period
• π is pie = 22/7
• L is pendulum length
• g is acceleration due to gravity

Given:

• T = 0.500 s

• g = 9.8 m/s²

solving step-wise:

$\dashrightarrow \mathrm{T}=2 \pi \sqrt{\dfrac{\mathrm{L}}{\mathrm{g}}}$

$\sf \dashrightarrow \mathrm{0.5}=2 \pi \sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}$

$\dashrightarrow \mathrm{\dfrac{0.5}{2 \pi } }=\sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}$

$\dashrightarrow\sqrt{\dfrac{\mathrm{L}}{\mathrm{9.8}}}= \mathrm{\dfrac{0.5}{2 \pi } }$

$\sf \dashrightarrow{\dfrac{\mathrm{L}}{\mathrm{9.8}}}= (\mathrm{\dfrac{0.5}{2 \pi } })^2$

$\sf \dashrightarrow{{\mathrm{L}}= (\mathrm{\dfrac{0.5}{2 \pi } })^2*9.8$

$\sf \dashrightarrow{{\mathrm{L}}=0.06205922 \ m$

• 1 m → 100 cm

$\sf \dashrightarrow{{\mathrm{L}}=6.2059\ cm$

$\sf \dashrightarrow{{\mathrm{L}}=6.21\ cm$             { rounded to nearest hundredth }

Answer 2

Let's see

$\\ \rm\rightarrowtail T=2\pi \sqrt{\dfrac{l}{g}}$

$\\ \rm\rightarrowtail 0.5=2\pi \sqrt{\dfrac{l}{9.8}}$

$\\ \rm\rightarrowtail 0.783=\pi \sqrt{l}$

$\\ \rm\rightarrowtail 0.2494=\sqrt{l}$

$\\ \rm\rightarrowtail \ell=0.0622m$