A Pendulum Taken to the Moon

Q. The acceleration due to gravity on the surface of moon is 1.7 m/s². What is the time period of a simple pendulum on the surface of moon if its time period on the surface of earth is 3.5 s, where g is 9.8 m/s², in seconds? expects 8.4 s

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Q. The acceleration due to gravity on the surface of moon is 1.7 m/s² $g_{m} = 1.7$ m/s². What is the time period of a simple pendulum $L_{l} = L_{p}$ on the surface of moon if its time period on the surface of earth is 3.5 s $T_{p} = 3.5$ s, where g is 9.8 m/s² $g_{e} = 9.8$ m/s², in seconds?

The playground

Inputs

moon g

g_{m}

1.7 m/s²

earth period

T_{p}

3.50 s

earth g

g_{e}

9.8 m/s²

Outputs

moon period

T_{l}

The solution

Solved with SimplePendulum

Given

$g_{m} = 1.7$ m/s²
$g_{e} = 9.8$ m/s²
$T_{p} = 3.5$ s
$L_{l} = L_{p}$ from “a simple pendulum”

Find $T_{l}$ .

Working

period of a simple pendulum(pendulum) $T_{p} = 2 \cdot π \cdot \sqrt{\frac{L_{p}}{g_{e}}}$
$3.5 = 2 \cdot 3.14159265 \cdot \sqrt{\frac{L_{p}}{9.8}}$
$L_{p} = 3.04090202$ m
a simple pendulum $L_{l} = L_{p}$
$L_{l} = 3.04090202$
$L_{l} = 3.04090202$ m
period of a simple pendulum(lunar pendulum) $T_{l} = 2 \cdot π \cdot \sqrt{\frac{L_{l}}{g_{m}}}$
$T_{l} = 2 \cdot 3.14159265 \cdot \sqrt{\frac{3.04090202}{1.7}}$
$T_{l} = 8.40 s$
$T_{l} \approx 8.4$ sto 2 significant figures

Conclusion

The same string swings in weaker gravity, so the bob is pulled back more gently and each swing takes longer, by the square root of how much weaker gravity is.

The Ruby

1require_relative "../physics"23question :a_pendulum_taken_to_the_moon do4  kind :numerical5  describes "The acceleration due to gravity on the surface of moon is %{moon_g}. %{moon_period} of a simple " \6            "pendulum on the surface of moon if its time period on the surface of earth is %{earth_period}, where g " \7            "is %{earth_g}, in seconds?"8  given moon_g: "1.7 m/s2"9  given earth_period: "3.5 s"10  given earth_g: "9.8 m/s2"11  asks moon_period: "What is the time period"12end

1require_relative "../questions/a_pendulum_taken_to_the_moon"2require_relative "../science/oscillations/simple_pendulum"34solve :a_pendulum_taken_to_the_moon, using: SimplePendulum do5  objects :earth, :moon6  object :pendulum, on: earth7  object :lunar_pendulum, on: moon89  take moon_g: moon.surface_gravity10  take earth_g: earth.surface_gravity11  take earth_period: pendulum.period1213  deduce_from "a simple pendulum", lunar_pendulum.length_of_the_string => pendulum.length_of_the_string1415  moon_period = lunar_pendulum.period16  answer moon_period:, in: "s", figures: 21718  conclusion "The same string swings in weaker gravity, so the bob is pulled back more gently and each swing " \19             "takes longer, by the square root of how much weaker gravity is."20end

1require_relative "../solutions/a_pendulum_taken_to_the_moon"23play_with :a_pendulum_taken_to_the_moon do4  title "A Pendulum Taken to the Moon"5end

The laws

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	—
period of lunar pendulum	$T_{l}$	s	—
length of the string of lunar pendulum	$L_{l}$	m	—
surface gravity of moon	$g_{m}$	m/s²	—
length of the string of earth	$L_{e}$	m	—
period of earth	$T_{e}$	s	—
length of the string of moon	$L_{m}$	m	—
period of moon	$T_{m}$	s	—
surface gravity of pendulum	$g_{p}$	m/s²	—
surface gravity of lunar pendulum	$g_{l}$	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)

T_{l} = 2 \cdot π \cdot \sqrt{\frac{L_{l}}{g_{m}}}

period of a simple pendulum(lunar pendulum)

L_{l} = L_{p}

a simple pendulum