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Chapter 16 - 16.19

19. (a) Eq. 16-8 leads to

ω =

= 123

0 . 100

a =

−

ω 2 x =

⇒

a

x

which yields ω =35 . 07 rad/s. Therefore, f = ω/ 2 π =5 . 58 Hz.

(b) Eq. 16-12 provides a relation between ω (found in the previous part) and the mass:

ω = k

m

= ⇒ m =

400

35 . 07 2

=0 . 325 kg .

(c) Byenergyconservation, 2 kx 2 m (the energy ofthe system at a turning point) is equal to the sum of

kinetic and potential energies at the time t described in the problem.

1

2 kx 2 m =

2 mv 2 + 1

2 kx 2

=

⇒

x m = m

k v 2 + x 2

.

Consequently, x m = (0 . 325 / 400)(13 . 6) 2 +0 . 1 2 =0 . 400 m.

−

1