P17_008 - 8. (a) The figure in the book makes it clear...

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Unformatted text preview: 8. (a) The figure in the book makes it clear that the period is T = 10 s and the amplitude is ym = 4.0 cm. The phase constant φ is more subtly determined by that figure: what is shown is 4 sin ωt, yet what follows from Eq. 17-2 (without the phase constant) should be 4 sin(−ωt) at x = 0. Thus, we need the phase constant φ = π since 4 sin(−ωt + π ) = 4 sin(ωt)). Therefore, we use Eq. 17-2 (modified by the inclusion of φ) with k = 2π/λ = π/10 (in inverse centimeters) and ω = 2π/T = π/5 (in inverse seconds). In the graph below we plot the equation for t = 0 over the range 0 ≤ x ≤ 20 cm, making sure our calculator is in radians mode. 4 y_centimeters 2 0 2 4 6 8 10 12 14 x_centimeters 16 18 20 –2 –4 (b) Since the frequency is f = 1/T = 0.10 s, the speed of the wave is v = f λ = 2.0 cm/s. (c) Using the observations made in part (a), Eq. 17-2 becomes y = 4.0 sin π x πt − +π 10 5 = −4.0 sin π x πt − 10 5 where y and x are in centimeters and t is in seconds. (d) Taking the derivative of y with respect to t, we find u= ∂y π π x πt = 4. 0 cos − ∂t t 10 5 which (evaluated at (x, t) = (0, 5.0), making sure our calculator is in radians mode) yields u = −2.5 cm/s. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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