Question

< Homework_chapter 11 Problem 11.45 - Enhanced - with Solution A certain simple pendulum has a period on earth of 1.71 s. You may want to review (Page) For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of the lost astronaut:

Answer 1

First, find the length of the simple pendulum.

Use formula $T=2\pi \sqrt{\frac{l}{g}}$

$\frac{T}{2\pi}= \sqrt{\frac{l}{g}}$

$(\frac{T}{2\pi})^{2}=\frac{l}{g}$

$\frac{T^{2}}{4\pi^{2}}*g=l$

$\frac{(1.71s)^{2}}{4\pi^{2}}*9.81m/s^{2}=l$

${\color{Red} l=0.7266m}$

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Time period on Mars, $T_{m}=2\pi \sqrt{\frac{l}{g_{m}}}$

$T_{m}=2\pi \sqrt{\frac{0.7266m}{3.71m/s^{2}}}$

ANSWER: ${\color{Red} T_{m}=2.781s}$

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