Question

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

Please show all of your steps to find the solution.

Answer 1

Let's see

`\\ \rm\rightarrowtail T=2\pi \sqrt{(l)/(g)}`

`\\ \rm\rightarrowtail 0.5=2\pi \sqrt{(l)/(9.8)}`

`\\ \rm\rightarrowtail 0.783=\pi √(l)`

`\\ \rm\rightarrowtail 0.2494=√(l)`

`\\ \rm\rightarrowtail \ell=0.0622m`

Answer 2

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{\frac{\mathrm{L}}{\mathrm{g}}}`

`\sf \dashrightarrow \mathrm{0.5}=2 \pi \sqrt{\frac{\mathrm{L}}{\mathrm{9.8}}}`

`\dashrightarrow \mathrm{(0.5)/(2 \pi ) }=\sqrt{\frac{\mathrm{L}}{\mathrm{9.8}}}`

`\dashrightarrow\sqrt{\frac{\mathrm{L}}{\mathrm{9.8}}}= \mathrm{(0.5)/(2 \pi ) }`

`\sf \dashrightarrow{\frac{\mathrm{L}}{\mathrm{9.8}}}= (\mathrm{(0.5)/(2 \pi ) })^2`

`\sf \dashrightarrow{{\mathrm{L}}= (\mathrm{(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 }