This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 647 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 CHAPTER 15 648 (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....
View
Full
Document
This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics, Acceleration

Click to edit the document details