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Unformatted text preview: 21. (a) We read the amplitude from the graph. It is about 5.0 cm.
(b) We read the wavelength from the graph. The curve crosses y = 0 at about x = 15 cm
and again with the same slope at about x = 55 cm, so
λ = (55 cm – 15 cm) = 40 cm = 0.40 m.
(c) The wave speed is v = τ / μ , where τ is the tension in the string and μ is the linear
mass density of the string. Thus,
v= 3.6 N
= 12 m/s.
25 ×10−3 kg/m (d) The frequency is f = v/λ = (12 m/s)/(0.40 m) = 30 Hz and the period is
T = 1/f = 1/(30 Hz) = 0.033 s.
(e) The maximum string speed is
um = ωym = 2πfym = 2π(30 Hz) (5.0 cm) = 940 cm/s = 9.4 m/s.
(f) The angular wave number is k = 2π/λ = 2π/(0.40 m) = 16 m–1.
(g) The angular frequency is ω = 2πf = 2π(30 Hz) = 1.9×102 rad/s
(h) According to the graph, the displacement at x = 0 and t = 0 is 4.0 × 10–2 m. The
formula for the displacement gives y(0, 0) = ym sin φ. We wish to select φ so that
5.0 × 10–2 sin φ = 4.0 × 10–2.
The solution is either 0.93 rad or 2.21 rad. In the first case the function has a positive
slope at x = 0 and matches the graph. In the second case it has negative slope and does
not match the graph. We select φ = 0.93 rad.
(i) The string displacement has the form y (x, t) = ym sin(kx + ωt + φ). A plus sign appears
in the argument of the trigonometric function because the wave is moving in the negative
x direction. Using the results obtained above, the expression for the displacement is
y ( x, t ) = ( 5.0 × 10−2 m ) sin (16 m −1 ) x + (190s −1 )t + 0.93 . ...
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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