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ISM_CH15

# ISM_CH15 - Chapter 15 1(a The amplitude is half the range...

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Chapter 15 1. (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 - × × 2. (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. 15-7) 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 (b) Using Eq. 15-12, we obtain ( 29 ( 29 2 2 0.12kg 10 rad/s 1.2 10 N/m. k k m ϖ π = = = × 3. (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 657

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CHAPTER 15 4. The textbook notes (in the discussion immediately after Eq. 15-7) 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 = 2 6.60 0.0220 = 37.8 2 2 a m π Hz m m / s b g c hb g . 5. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = 1/ T = 1/(0.500 s) = 2.00 Hz. (c) The angular frequency ϖ is ϖ = 2 π f = 2 π (2.00 Hz) = 12.6 rad/s. (d) The angular frequency is related to the spring constant k and the mass m by ϖ = k m . We solve for k : k = m ϖ 2 = (0.500 kg)(12.6 rad/s) 2 = 79.0 N/m. (e) Let x m be the amplitude. The maximum speed is v m = ϖ x m = (12.6 rad/s)(0.350 m) = 4.40 m/s. (f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by F m = kx m = (79.0 N/m)(0.350 m) = 27.6 N. 6. (a) The problem describes the time taken to execute one cycle of the motion. The period is T = 0.75 s. (b) Frequency is simply the reciprocal of the period: f = 1/ T 1.3 Hz, where the SI unit abbreviation Hz stands for Hertz, which means a cycle-per-second. (c) Since 2 π radians are equivalent to a cycle, the angular frequency ϖ (in radians-per- second) is related to frequency f by ϖ = 2 π f so that ϖ 8.4 rad/s. 7. (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.
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