ch16-p009 - 9. (a) Recalling from Ch. 12 the simple...

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Unformatted text preview: 9. (a) Recalling from Ch. 12 the simple harmonic motion relation um = ymω, we have ω= 16 = 400 rad/s. 0.040 Since ω = 2πf, we obtain f = 64 Hz. (b) Using v = fλ, we find λ = 80/64 = 1.26 m ≈ 1.3 m . (c) The amplitude of the transverse displacement is ym = 4.0 cm = 4.0 ×10−2 m. (d) The wave number is k = 2π/λ = 5.0 rad/m. (e) The angular frequency, as obtained in part (a), is ω = 16 / 0.040 = 4.0 ×102 rad/s. (f) The function describing the wave can be written as y = 0.040sin ( 5 x − 400t + φ ) where distances are in meters and time is in seconds. We adjust the phase constant φ to satisfy the condition y = 0.040 at x = t = 0. Therefore, sin φ = 1, for which the “simplest” root is φ = π/2. Consequently, the answer is y = 0.040sin 5 x − 400t + (g) The sign in front of ω is minus. π . 2 ...
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