Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 4 - Graphs of the Circular Functions - Section 4.5 Harmonic Motion - 4.5 Exercises - Page 185: 13

Answer

$P=\frac{\pi}{4}$ seconds $T=\frac{4}{\pi}$ oscillations/second

Work Step by Step

Since we know that the length is 0.5 ft, we can use the following formula to find the period P: $P=2\pi\sqrt {\frac{L}{32}}$ $P=2\pi\sqrt {\frac{0.5}{32}}$ $P=2\pi\sqrt {\frac{1}{64}}$ $P=2\pi(\frac{1}{8})$ $P=\frac{\pi}{4}$ seconds The frequency T is a reciprocal of period P. Therefore, $T=\frac{1}{P}=\frac{1}{\frac{\pi}{4}}=\frac{4}{\pi}$ oscillations/second
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