A Pendulum That Ticks Seconds

solved SimplePendulum

Q. Taking g as 9.8 m/s² $g_{e} = 9.8$ m/s², what is the length of a simple pendulum, which ticks seconds $T_{p} = 2$ s, in metres?

Inputs

gravity

g_{e}

9.8 m/s²

Outputs

length

L_{p}

How it's solved

Solved with SimplePendulum

Given

$g_{e} = 9.8$ m/s²
$T_{p} = 2$ s from “ticks seconds”

Find $L_{p}$ .

Working

period of a simple pendulum(pendulum) $T_{p} = 2 \cdot π \cdot \sqrt{\frac{L_{p}}{g_{e}}}$
$2 = 2 \cdot 3.14159265 \cdot \sqrt{\frac{L_{p}}{9.8}}$
$L_{p} = 0.99 m$
$L_{p} \approx 0.99$ mto 2 significant figures

Conclusion

A pendulum that ticks once a second swings there and back in two, and that period fixes its length: about a metre.

The web of concepts

given read from the words equation asked

The Ruby that solves it

The question and its solution are small Ruby files. The engine runs them.

question.rb

1require_relative "../physics"23question :a_pendulum_that_ticks_seconds do4  kind :numerical5  describes "Taking g as %{gravity}, %{length}, in metres?"6  given gravity: "9.8 m/s2"7  asks length: "what is the length of a simple pendulum, which ticks seconds"8end

solution.rb

1require_relative "../questions/a_pendulum_that_ticks_seconds"2require_relative "../science/oscillations/simple_pendulum"34solve :a_pendulum_that_ticks_seconds, using: SimplePendulum do5  objects :earth6  object :pendulum, on: earth78  take gravity: earth.surface_gravity910  deduce_from "ticks seconds", pendulum.period => "2 s"1112  length = pendulum.length_of_the_string13  answer length:, in: "m", figures: 21415  conclusion "A pendulum that ticks once a second swings there and back in two, and that period fixes " \16             "its length: about a metre."17end

playground.rb

1require_relative "../solutions/a_pendulum_that_ticks_seconds"23play_with :a_pendulum_that_ticks_seconds do4  title "A Pendulum That Ticks Seconds"5end

The laws it drew on

quantity	written	units	within	from
pi	$π$	—	0 … ∞	SimplePendulum
surface gravity	$g$	m/s²	—	SimplePendulum
length of the string	$L$	m	—	SimplePendulum
period	$T$	s	—	SimplePendulum
period of pendulum	$T_{p}$	s	—
length of the string of pendulum	$L_{p}$	m	—
surface gravity of earth	$g_{e}$	m/s²	—
length of the string of earth	$L_{e}$	m	—
period of earth	$T_{e}$	s	—
surface gravity of pendulum	$g_{p}$	m/s²	—

T_{b} = 2 \cdot π \cdot \sqrt{\frac{L_{b}}{g_{g}}}

period of a simple pendulum

T_{p} = 2 \cdot π \cdot \sqrt{\frac{L_{p}}{g_{e}}}

period of a simple pendulum(pendulum)