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 186: 16a


Amplitude= 0.5 Period$=4.44$ seconds Frequency$=0.225$ cycles per second

Work Step by Step

In the equation $s(t)=a\sin \sqrt \frac{k}{m}t$, we substitute the values of $k$ and $m$: $s(t)=a\sin \sqrt \frac{k}{m}t$ $s(t)=a\sin \sqrt \frac{2}{1}t$ $s(t)=a\sin \sqrt{2} t$ Since we know that the spring is stretched 0.5 ft and then released, this means that the amplitude is 0.5. Next, to find the period, we need to compare the equation above with the standard equation $s(t)=a\sin w t$. Comparing the two equations, we find that $w=\sqrt 2$. Therefore, the period can be found through the formula: Period$=\frac{2\pi}{w}$ Period$=\frac{2\pi}{\sqrt 2}$ Period$=4.44$ seconds Frequency is the reciprocal of period. Therefore, Frequency$=\frac{1}{4.44}=0.225$ cycles per second
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