Halliday 2 edicao

1. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = 2xm, where  is the angular frequency ( = 2f since there are 2 radians in one cycle). Therefore, in this circumstance, we obtain

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 = xm, where  is the angular frequency. Since  = 2f, where f is the frequency,

(c) The maximum acceleration is

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 = –xm is 2xm. Thus, xm = 36/2 = 18 cm.

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  is  = 2f = 2(2.00 Hz) = 12.6 rad/s.

(d) The angular frequency is related to the spring constant k and the mass m by . We solve for k and obtain

k = m2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.

(e) Let xm be the amplitude. The maximum speed is

vm = xm = (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.

11. (a) Making sure our calculator is in radians mode, we find

(b) Differentiating with respect to time and evaluating at t = 2.0 s, we find

(c) Differentiating again, we obtain

(d) In the