Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

Published by Pearson
ISBN 10: 0133942651
ISBN 13: 978-0-13394-265-1

Chapter 15 - Oscillations - Exercises and Problems: 65

Answer

$g = 5.87~m/s^2$

Work Step by Step

We can find the period of the oscillations as: $T = \frac{time}{oscillations}$ $T = \frac{14.5~s}{10~oscillations}$ $T = 1.45~s$ We can use the period to find the spring constant as: $T= 2\pi~\sqrt{\frac{m}{k}}$ $k = \frac{(2\pi)^2~m}{T^2}$ $k = \frac{(2\pi)^2~(0.200~kg)}{(1.45~s)^2}$ $k = 3.76~N/m$ When the spring is stretched 31.2 cm, the force of the spring is equal to the weight of the mass. We then find $g$: $mg = kx$ $g = \frac{kx}{m}$ $g = \frac{(3.76~N/m)(0.312~m)}{0.200~kg}$ $g = 5.87~m/s^2$
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