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- Date Submitted: 09/24/2013 6:17 PM
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Chapter 15

1. (a) The amplitude is half the range of the displacement, or xm = 1.0 mm.

(b) The maximum speed vm is related to the amplitude xm by vm = ωxm, where ω is the angular frequency. Since ω = 2πf, where f is the frequency,

[pic]

(c) The maximum acceleration is

[pic]

2. (a) The acceleration amplitude is related to the maximum force by Newton’s second law: Fmax = mam. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency (ω = 2πf since there are 2π 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 ω = 10π (understood to be valid to two significant figures). Therefore,

[pic]

(b) Using Eq. 15-12, we obtain

[pic]

3. (a) The angular frequency ω is given by ω = 2πf = 2π/T, where f is the frequency and T is the period. The relationship f = 1/T was used to obtain the last form. Thus

ω = 2π/(1.00 × 10–5 s) = 6.28 × 105 rad/s.

(b) The maximum speed vm and maximum displacement xm are related by vm = ωxm, so

[pic]

4. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency (ω = 2πf since there are 2π radians in one cycle). Therefore, in this circumstance, we obtain

[pic]

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 ω is ω = 2πf = 2π(2.00 Hz) = 12.6 rad/s.

(d) The angular frequency is related to the spring constant k and the mass m by [pic]. We solve for k: k = mω2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.

(e) Let xm be the amplitude. The maximum speed is vm = ωxm = (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 Fm = kxm = (79.0...