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
.
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Simple Harmonic Motion

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