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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, vm = 2π fxm = 2π (120 Hz ) (1.0 ×10 −3 m ) = 0.75 m/s.

(c) The maximum acceleration is am = ω 2 xm = ( 2π f ) xm = ( 2π (120 Hz ) ) (1.0 ×10−3 m ) = 5.7 ×102 m/s 2 .
2 2

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, Fmax = mω 2 xm = 0.12 kg 10π rad / s (b) Using Eq. 15-12, we obtain

b

gb

g b0.085 mg = 10 N.
2
2

ω=

k m

k = ( 0.12kg )(10π rad/s ) = 1.2 ×102 N/m.

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 1.00 × 103 m / s xm = = = 1.59 × 10−3 m. 5 ω 6.28 × 10 rad / s vm

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

am = 2 π 6.60 Hz

c b

gh b0.0220 mg = 37.8 m / s .
2

2

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