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
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/Twas 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
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 thefrequency,
(c) The maximum acceleration is
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 theperiod: 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,
(b) Using Eq. 15-12, we obtain
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 = 0.25second 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 =–xm is 2xm. Thus, xm = 36/2 = 18 cm.
6. (a) Since the problem gives the frequency f = 3.00 Hz, we have ω = 2πf = 6π rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of the mass mcar so that Eq. 15-12 leads to
(b) If the new mass being supported by the four springs is mtotal = [1450 + 5(73)] kg = 1815 kg, then Eq. 15-12 leadsto
7. (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 and obtain
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 N/m)(0.350 m) = 27.6 N.
8. (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 thereciprocal 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 2π radians are equivalent to a cycle, the angular frequency ω (in radians-per-second) is related to frequency f by ω = 2πf so that ω ≈ 8.4 rad/s.
9. The magnitude of the maximum acceleration is given by am = ω2xm, where ω is the angular frequency and xm is theamplitude.
(a) The angular frequency for which the maximum acceleration is g is given by [pic], and the corresponding frequency is given by
(b) For frequencies greater than 498 Hz, the acceleration exceeds g for some part of the motion.
10. We note (from the graph) that xm = 6.00 cm. Also the value at t = 0 is xo = − 2.00 cm. Then Eq. 15-3 leads to
φ = cos−1(−2.00/6.00)...
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