P16_020.PDF

20. Eq. 16-12 gives the angular velocity:

ω = k

m = 100

2 . 00 =7 . 07 rad / s .

16-4) provide one method of solution. Here, we use trigonometric tech-

niques based on Eq. 16-3 and Eq. 16-6.

(a) Dividing Eq. 16-6 by Eq. 16-3, we obtain

x =

−

ω tan( ωt + φ )

so that the phase ( ωt + φ ) is found from

ωt + φ =tan − 1 −

ωx

=tan − 1

3 . 415

(7 . 07)(0 . 129)

−

With the calculator in radians mode, this gives the phase equal to

1 . 31 rad. Plugging this back

into Eq. 16-3 leads to

0 . 129m = x m cos( − 1 . 31) = ⇒ 0 . 500m = x m .

1 . 31 rad at t =1 . 00 s. We can use the above value of ω to solve for the phase

constant φ .Weobtain φ =

−

8 . 38 rad (though this, as well as the previous result, can have 2 π or

4 π (and so on) added to it without changing the physics of the situation). With this value of φ ,we

ﬁnd x o = x m cos φ =

−

0 . 251 m.

−

x m ω sin φ =3 . 06 m/s.

Energy methods (discussed in

−

(b) Since ωt + φ =