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HW11_Sol_F10

# HW11_Sol_F10 - Nov 8 2010 PHYS 2101 HW#11 WileyPlus Problem...

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Nov. 8, 2010 PHYS 2101 HW#11 WileyPlus Problem Solutions 1 HW #11 Solutions – or at least most of them… (USING BOOK VALUES 9 th Edition in problems; problem # indicated by []) 1) [16-CQ4] 2) [16-3] . (a) The angular wave number is 1 2 2 3.49m . 1.80m k π π = = = λ (b) The speed of the wave is ( ) ( ) 1.80m 110rad s 31.5m s. 2 2 v f ω λ = λ = = = π π 3) [16-5] (a) The motion from maximum displacement to zero is one-fourth of a cycle. One-fourth of a period is 0.170 s, so the period is T = 4(0.170 s) = 0.680 s. (b) The frequency is the reciprocal of the period: 1 1 1.47Hz. 0.680s f T = = = (c) A sinusoidal wave travels one wavelength in one period: 1.40m 2.06m s. 0.680s v T = = = λ 4) [16-9] (a) The amplitude y m is half of the 6.00 mm vertical range shown in the figure, that is, 3.0 mm. m y = (b) The speed of the wave is v = d/t = 15 m/s, where d = 0.060 m and t = 0.0040 s. The angular wave number is k = 2 π / λ where λ = 0.40 m. Thus, k = 2 π λ = 16 rad/m . (c) The angular frequency is found from ω = k v = (16 rad/m)(15 m/s) = 2.4 × 10 2 rad/s. (d) We choose the minus sign (between kx and ω t ) in the argument of the sine function because the wave is shown traveling to the right ( in the + x direction, see Section 16-5). Therefore, with SI units understood, we obtain y = y m sin( kx kvt ) 0.0030 sin(16 x 2.4 × 10 2 t ) . 5) [16-10] (a) The amplitude is y m = 6.0 cm. (b) We find λ from 2 π / λ = 0.020 π : λ = 1.0 × 10 2 cm. (c) Solving 2 π f = ω = 4.0 π , we obtain f = 2.0 Hz. (d) The wave speed is v = λ f = (100 cm) (2.0 Hz) = 2.0 × 10 2 cm/s. (e) The wave propagates in the – x direction, since the argument of the trig function is kx + ω t instead of kx ω t (as in Eq. 16-2).

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HW11_Sol_F10 - Nov 8 2010 PHYS 2101 HW#11 WileyPlus Problem...

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