University Physics with Modern Physics (14th Edition)

Published by Pearson
ISBN 10: 0321973615
ISBN 13: 978-0-32197-361-0

Chapter 14 - Periodic Motion - Problems - Exercises - Page 459: 14.1

Answer

a.) Music. T = 2.15 x 10^{-3} s \omega = 2.93 x 10^(3) rad/s b.) Hearing. f = 2.0 x 10^{4} Hz \omega = 1.26 x 10^{5} rad/s c.) Vision. f = 4.3 x 10^{14} Hz to f = 7.5 x 10^{14} Hz T = 1.3 x 10^{-15} s to T = 2.5 x 10^{-15} s d.) Ultrasound. T = 2.0 x 10^{-7} s \omega = 3.1 x 10^{7} rad/s

Work Step by Step

a.) Music. required: T and \omega given: f = 466 Hz Solution: The period T and frequency f are related by the equation: T = \frac{1}{f} Therefore the period that it takes for the vocal chords to vibrate through one complete cycle is T = \frac{1}{466hz} = 2.15 x 10^{-3} s. The angular frequency is given by the equation: \omega = 2\pi f Therefore the angular frequency of the chords \omega = 2\pi (466Hz) = 2.93 x 10^(3) rad/s. b.) Hearing. required: f and \omega given: 50.0 \mu s Solution: First, we convert 50.0 \mu s into SI units, which gives: 5 x 10^{-5} s. The period T and frequency f are related by the equation: T = \frac{1}{f}. We can multiply both sides of the equation by \frac{f}{T} so we get the equation for the frequency: f = \frac{1}{T} Solving for the frequency we get, f = \frac{1}{T} = \frac{1}{5 x 10^{-5}} = 2.0 x 10^{4} Hz The angular frequency is given by the equation: \omega = 2\pi f Therefore the angular frequency of the vibrating eardrum is \omega = 2\pi f = 2\pi (2.0 x 10^{4} Hz = 1.26 x 10^{5} rad/s. c.) Vision. required: Limits of the period T and frequency F given: \omega = 2.7 x 10^{15} rad/s and \omega = 4.7 x 10^{15} rad/s Solution: The frequency f and angular frequency \omega is related by \omega = 2\pi f , and by dividing both sides by 2\pi we get f = \frac{\omega}{2\pi}. The lower limit is f = \frac{2.7 x 10^{15} rad/s}{2\pi rad} = 4.3 x 10^{14} Hz. The upper limit is f = \frac{4.7 x 10^{15} rad/s}{2\pi rad} = 7.5 x 10^{14} Hz. The period T and frequency f is related by: T = \frac{1}{f}. The lower limit for the T is T = \frac{1}{7.5 x 10^{14} Hz} = 1.3 x 10^{-15} s. And the upper limit for the T is T = \frac{1}{4.3 x 10^{14} Hz } = 2.5 x 10^{-15} s. d.) Ultrasound. required: T and \omega given: 5.0 MHz Solution: First, we convert 5.0 MHz to SI units, which gives: 5 x 10^{6} Hz. The period T and frequency f is related by T = \frac{1}{f} Therefore, T = \frac{1}{5 x 10^{6} Hz} = 2.0 x 10^{-7} s. The frequency and angular frequency \omega is related by \omega = 2\pi f Therefore, \omega = 2\pi (5 x 10^{6} Hz) = 3.1 x 10^{7} rad/s.
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