Arte

Chapter 15

1. 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

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2. (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

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3. (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,

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(c) The maximum acceleration is

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4. (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,

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

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5. (a) During simple harmonic motion, the speed is (momentarily) zero when the object is at a “turning point” (that is, when x = +xm or x = –xm). Consider that it starts at x = +xm and we are told that t = 0.25 second elapses until the object reaches x = –xm. To execute a full cycle of the motion (which takes a period T to complete), the object which started at x = +xm must return to x = +xm (which, by symmetry, will occur 0.25 second after it was at x = –xm). Thus, T = 2t = 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 = +xm and x =