Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

Published by Pearson
ISBN 10: 0321740904
ISBN 13: 978-0-32174-090-8

Chapter 14 - Oscillations - Exercises and Problems: 2

Answer

(a) 3.3 s (b) 0.30 Hz (c) 1.9 rad/s (d) A=25 cm (e) 48 cm/s

Work Step by Step

(a) The period is time taken to complete one cycle. This is $$\frac{33s}{10 oscillations}=3.3s/oscillation$$ Therefore, $T=3.3s$. (b) The frequency is equal to $$f=\frac{1}{T}$$ Substituting the value of $T=3.3s$ yields a frequency of $$f=\frac{1}{3.3s}=0.30Hz$$ (c) The angular frequency $\omega$ equals $\omega=2\pi f$. Substituting $f=0.30Hz$ yields $\omega=2\pi(0.30Hz)=1.9rad/s$ (d) The amplitude is the distance from equilibrium to maximum compression. Since the minimum and maximum distance are separated by $50cm$, the amplitude must be half of this value, $25cm$. (e) The maximum speed of an oscillator equals $v_m=A\omega$. Substituting known values of $A=25cm$ and $\omega=1.9rad/s$ yields a maximum speed of $$v_m=(25cm)(1.9rad/s)=48cm/s$$
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