Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 11 - Oscillations and Waves - General Problems - Page 325: 68

Answer

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Work Step by Step

The equation of motion is given as $x=0.25sin(4.70t)=A\;sin\omega t$. a. The amplitude is the maximum x value, or 0.25 meters. b. The frequency $f = \frac{\omega}{2 \pi}=\frac{4.70\;rad/s}{2 \pi}=0.748\;Hz$ c. The period $T =\frac{2 \pi}{\omega}=\frac{2 \pi}{4.70\;rad/s}=1.34\;s$ d. The total energy is equal to the kinetic energy at maximum speed. $$E_T=\frac{1}{2}mv_{max}^2=\frac{1}{2}m(\omega\;A)^2=\frac{1}{2}(0.650\;kg)((4.70\;rad/s)\;0.25\;m)^2\approx 0.45\;J$$ e. First find the potential energy when x = 15 cm. $$PE=\frac{1}{2}kx^2=\frac{1}{2}m\omega^2x^2=\frac{1}{2}(0.650\;kg)(4.70\;rad/s)^2(0.15\;m)^2= 0.1615\;J$$ The kinetic energy is the total energy minus the potential energy. $$KE=0.4487\;J-0.1615\;J\approx 0.29\;J$$
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