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Chapter 17 - 17.70

70. (a) Recalling from Ch . 12 the simple harmonic motion relation u m = y m ω ,wehave

ω =

16

0 . 04 = 400 rad / s .

Since ω =2 πf ,weobtain f =64Hz.

(b) Using v = fλ , we ﬁnd λ =80 / 64 = 1 . 26 m.

(c) Now, k =2 π/λ = 5 rad/m, so the function describing the wave becomes

y =0 . 04sin(5 x

−

400 t + φ )

where distances are in meters and time is in seconds. We adjust the phase constant φ to satisfy the

condition y =0 . 04 at x = t = 0. Therefore, sin φ = 1, for which the “simplest” root is φ = π/ 2.

Consequently, the answer is

400 t + π

2

.

−

y =0 . 04sin 5 x