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 326: 74

Answer

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Work Step by Step

a. Equation 11-5b relates the velocity and the position of a simple harmonic oscillator. Set v equal to half the maximum speed, and solve for the displacement. $$v=v_{max}\sqrt{1-x^2/x_0^2}=\frac{1}{2}v_{max}$$ $$ \sqrt{1-x^2/x_0^2}=\frac{1}{2} $$ $$ 1-x^2/x_0^2=\frac{1}{4} $$ $$ x^2/x_0^2=\frac{3}{4} $$ $$x=\frac{\sqrt{3}}{2}x_0$$ b. For a mass on a spring, the magnitude of the force is proportional to the distance from equilibrium, F = -kx, so the acceleration is also proportional to the displacement. This may be expressed as $a=-kx/m$. Therefore, the acceleration reaches half its maximum value when the displacement also reaches half its maximum value. The distance is $x=\frac{1}{2}x_0$.
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