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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 is the angular frequency. Since = 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-× (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-× × 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 = 2 x m , where is the angular frequency ( = 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 = 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 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 = ⇒ = = × 3. (a) The angular frequency is given by = 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 = 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 = x m , so = = 1.00 10 6.28 10 = 1.59 10 . 3 5 3 x v m m × × ×-m / s rad / s m 657 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 is the angular frequency ( = 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 is = 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 = k m . We solve for k : k = m 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 = 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 (in radians-per-second) is related to frequency f by = 2 π f so that ≈ 8.4 rad/s....
View Full Document

This note was uploaded on 12/17/2009 for the course PHY 108 taught by Professor Iashvili during the Spring '08 term at SUNY Buffalo.

Page1 / 46

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online