How long must a pendulum be the Moon, where g-16 N/kg, to have period of 2.0 s?

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Answer 1 · 2018-06-11T12:51:53

I get approximately #0.162# meters or #16.2# centimeters.

The period of a simple pendulum is given by the equation:

#T=2pisqrt(l/g)#

where:

#T# is the period of the pendulum in seconds

#l# is the length of the pendulum in meters

#g# is the gravitational acceleration

I'm assuming that the gravitational acceleration on the moon is taken as #1.6 \ "m/s"^2#, since it must be weaker than Earth's.

So, we get:

#2 \ "s"=2pisqrt(l/(1.6 \ "N/kg"))#

#2 \ "s"=2pisqrt(l/(1.6 \ "m/s"^2))#

#(2 \ "s")/(2pi)=sqrt(l/(1.6 \ "m/s"^2))#

#"s"/pi=sqrt(l/(1.6 \ "m/s"^2))#

#"s"^2/pi^2=l/(1.6 \ "m/s"^2)#

#l=(color(red)cancelcolor(black)("s"^2))/pi^2*(1.6 \ "m")/(color(red)cancelcolor(black)("s"^2))#

#l=(1.6 \ "m")/pi^2#

#~~0.162 \ "m"#