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 - Page 415: 1

Answer

(a) T = 3.3 s (b) f = 0.30 Hz (c) The amplitude is 25 cm (d) $v_{max} = 0.48~m/s$

Work Step by Step

(a) We can find the period as: $T = \frac{time}{oscillations}$ $T = \frac{33~s}{10~oscillations}$ $T = 3.3~s$ (b) We can find the frequency as: $f = \frac{1}{T}$ $f = \frac{1}{3.3~s}$ $f = 0.30~Hz$ (c) The glider oscillates between the 10-cm mark and the 60-cm mark. The equilibrium point is the midpoint which is 35 cm. The glider oscillates 25 cm on either side of the equilibrium point. Therefore, the amplitude is 25 cm. (d) We can find the maximum speed as: $v_{max} = \frac{2\pi~A}{T}$ $v_{max} = \frac{(2\pi)~(0.25~m)}{3.3~s}$ $v_{max} = 0.48~m/s$
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