hw13_sol_s11

hw13_sol_s11 - Phys 2101 Homework 13 Solution Spring `11...

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Phys 2101 Homework 13 Solution Spring `11 These solutions use the parameter values from the problems printed in the book, not those that appear in your personal homework assignment. The logic used to get to the answer is the same, however. 1. (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: 11 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 2. (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  where = 0.40 m. Thus, k = 2 = 16 rad/m . (c) The angular frequency is found from = k v = (16 rad/m)(15 m/s) = 2.4×10 2 rad/s. (d) We choose the minus sign (between kx and 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 ) . 3. (a) The amplitude is y m = 6.0 cm. (b) We find from 2 / = 0.020 : = 1.0×10 2 cm. (c) Solving 2 f = = 4.0 , we obtain f = 2.0 Hz.

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(d) The wave speed is v = 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 + t instead of kx t (as in Eq. 16-2). (f) The maximum transverse speed (found from the time derivative of y ) is    1 max 2 4.0 s 6.0cm 75cm s. m u fy   (g) y (3.5 cm, 0.26 s) = (6.0 cm) sin[0.020 (3.5) + 4.0 (0.26)] = 2.0 cm. 4. (a) The wave speed is given by v = / T = / k , where is the wavelength, T is the period, is the angular frequency (2 / T ), and k is the angular wave number (2 / ). The displacement has the form y = y m sin( kx + t ), so k = 2.0 m 1 and = 30 rad/s. Thus v = (30 rad/s)/(2.0 m 1 ) = 15 m/s. (b) Since the wave speed is given by v =  , where is the tension in the string and is the linear mass density of the string, the tension is     2 24 1.6 10 kg m 15m s 0.036N. v 5. The pulses have the same speed v . Suppose one pulse starts from the left end of the wire at time t = 0. Its coordinate at time t is x 1 = vt . The other pulse starts from the right end, at x = L , where L is the length of the wire, at time t = 30 ms. If this time is denoted by t 0 , then the coordinate of this wave at time t is x 2 = L
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This note was uploaded on 05/09/2011 for the course PHYS 2101 taught by Professor Grouptest during the Spring '07 term at LSU.

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hw13_sol_s11 - Phys 2101 Homework 13 Solution Spring `11...

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