P16_012.PDF

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Chapter 16 - 16.12
12. (a) Making sure our calculator is in radians mode, we find
x =6 . 0cos 3 π (2 . 0) + π
3
=3 . 0m .
(b) Differentiating with respect to time and evaluating at t =2 . 0 s, we find
v = dx
dt =
−
3 π (6 . 0)sin 3 π (2 . 0) + π
3
=
49 m / s .
(c) Differentiating again, we obtain
a = dv
−
(3 π ) 2 (6 . 0)cos 3 π (2 . 0) + π
3
=
−
2 . 7
×
10 2 m / s 2 .
20 rad.
(e) Comparing with Eq. 16-3, we see that ω =3 π rad/s. Therefore, f = ω/ 2 π =1 . 5Hz.
(f) The period is the reciprocal of the frequency: T =1 /f
≈
0 . 67 s.
−
dt =
(d) In the second paragraph after Eq. 16-3, the textbook defines the phase of the motion. In this case
(with t =2 . 0 s) the phase is 3 π (2 . 0) + 3 ≈

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