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42 votes
42 votes
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.

User Rockwell
by
3.1k points

2 Answers

9 votes
9 votes

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

User Guillim
by
2.6k points
9 votes
9 votes

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 }

User AKoran
by
3.2k points