Question

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 45.0 cm . The explorer finds that the pendulum completes 106 full swing cycles in a time of 129 s .What is the magnitude of the gravitational acceleration on this planet?

Answer 1

Given that the pendulum completes 106 full swing cycles in 129s

Time Period of a simple pendulum is the time taken to complete 1 full oscillation.

Time Period = Total time/No of oscillations

$T=\frac{129s}{106}$

${\color{Red} T=1.217s}$

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

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

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

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

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

$g=\frac{0.45m*4\pi^{2} }{(1.217s)^{2}}$

$g=\frac{0.45*4\pi^{2} }{1.217^{2}}$

ANSWER: ${\color{Red} g=12m/s^{2}}$

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