Physics (10th Edition)

Published by Wiley
ISBN 10: 1118486897
ISBN 13: 978-1-11848-689-4

Chapter 10 - Simple Harmonic Motion and Elasticity - Problems - Page 277: 44

Answer

The acceleration due to gravity on this planet is $6.04m/s^2$

Work Step by Step

In a simple pendulum, $$\frac{2\pi}{T}=\sqrt{\frac{g}{L}}$$ $$\frac{g}{L}=\frac{4\pi^2}{T^2}$$ $$g=\frac{4\pi^2L}{T^2}$$ Here, the pendulum's length $L=1.2m$. The pendulum finishes $100$ vibrations in $280s$, so the motion's period $T=\frac{280s}{100rev}=2.8s$ So the acceleration due to gravity on this planet is $$g=6.04m/s^2$$
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