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Unformatted text preview: Chapter 15 1. The textbook notes (in the discussion immediately after Eq. 157) that the acceleration amplitude is a m = ϖ 2 x m , where ϖ is the angular frequency ( ϖ = 2 π f since there are 2 π radians in one cycle). Therefore, in this circumstance, we obtain ( 29 ( 29 ( 29 2 2 2 2 (2 ) 2 6.60 Hz 0.0220 m 37.8 m/s . m m m a x f x ϖ π π = = = = 2. (a) The angular frequency ϖ is given by ϖ = 2 π f = 2 π / T , where f is the frequency and T is the period. The relationship f = 1/ T was used to obtain the last form. Thus ϖ = 2 π /(1.00 × 10 – 5 s) = 6.28 × 10 5 rad/s. (b) The maximum speed v m and maximum displacement x m are related by v m = ϖ x m , so = = 1.00 10 6.28 10 = 1.59 10 . 3 5 3 x v m m ϖ × × × m / s rad / s m 3. (a) The amplitude is half the range of the displacement, or x m = 1.0 mm. (b) The maximum speed v m is related to the amplitude x m by v m = ϖ x m , where ϖ is the angular frequency. Since ϖ = 2 π f , where f is the frequency, ( 29 ( 29 3 = 2 = 2 120 Hz 1.0 10 m = 0.75 m/s. m m v fx π π × (c) The maximum acceleration is ( 29 ( 29 ( 29 ( 29 2 2 2 3 2 2 = = 2 = 2 120 Hz 1.0 10 m = 5.7 10 m/s . m m m a x f x ϖ π π × × 4. (a) The acceleration amplitude is related to the maximum force by Newton’s second law: F max = ma m . The textbook notes (in the discussion immediately after Eq. 157) that the acceleration amplitude is a m = ϖ 2 x m , where ϖ is the angular frequency ( ϖ = 2 π f since there are 2 π radians in one cycle). The frequency is the reciprocal of the period: f = 1/ T = 1/0.20 = 5.0 Hz, so the angular frequency is ϖ = 10 π (understood to be valid to two significant figures). Therefore, = = 0.12 10 0.085 = 10 . 2 2 F m x m max kg rad / s m N ϖ b gb gb g π 647 CHAPTER 15 (b) Using Eq. 1512, we obtain ( 29 ( 29 2 2 2 0.12kg 10 rad/s 1.2 10 N/m. k k m m ϖ ϖ π = ⇒ = = = × 5. (a) During simple harmonic motion, the speed is (momentarily) zero when the object is at a “turning point” (that is, when x = + x m or x = – x m ). Consider that it starts at x = + x m and we are told that t = 0.25 second elapses until the object reaches x = – x m . To execute a full cycle of the motion (which takes a period T to complete), the object which started at x = + x m must return to x = + x m (which, by symmetry, will occur 0.25 second after it was at x = – x m ). Thus, T = 2 t = 0.50 s. (b) Frequency is simply the reciprocal of the period: f = 1/ T = 2.0 Hz. (c) The 36 cm distance between x = + x m and x = – x m is 2 x m . Thus, x m = 36/2 = 18 cm. 6. (a) Since the problem gives the frequency f = 3.00 Hz, we have ϖ = 2 π f = 6 π rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of the mass m car so that Eq. 1512 leads to ( 29 ( 29 2 5 car 1 1450kg 6 rad/s 1.29 10 N/m....
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This note was uploaded on 10/05/2010 for the course BIOL biol111 taught by Professor Ciccarello during the Spring '10 term at University of Massachusetts Boston.
 Spring '10
 Ciccarello
 Biology

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