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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
5 = −4.0 sin π x πt
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
π x πt
= 4. 0
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.
- Fall '08