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Unformatted text preview: 13. The wave speed v is given by v = , where is the tension in the rope and is the linear mass density of the rope. The linear mass density is the mass per unit length of rope: = m/L = (0.0 60 0 kg)/(2.00 m) = 0.0300 kg/m. Thus 500 N 129m s. 0.0300kg m v = = 15. (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 ( 29 ( 29 2 2 4 1.6 10 kg m 15m s 0.036 N. v - = = = 18. (a) Comparing with Eq. 16-2, we see that k = 20/m and = 600/s. Therefore, the speed of the wave is (see Eq. 16-13) v = / k = 30 m/s. (b) From Eq. 1626, we find 2 2 15 0.017kg m 17g m. 30 v = = = = 20. (a) The general expression for y ( x, t ) for the wave is y ( x, t ) = y m sin( kx t ), which, at x = 10 cm, becomes y ( x = 10 cm, t ) = y m sin[ k (10 cm t )]. Comparing this with the expression given, we find = 4.0 rad/s, or f = /2 = 0.64 Hz. (b) Since k (10 cm) = 1.0, the wave number is k = 0.10/cm. Consequently, the wavelength is = 2 / k = 63 cm. (c) The amplitude is 5.0 cm. m y = (d) In part (b), we have shown that the angular wave number is k = 0.10/cm. (e) The angular frequency is = 4.0 rad/s. (f) The sign is minus since the wave is traveling in the + x direction. Summarizing the results obtained above by substituting the values of k and into the general expression for y ( x, t ), with centimeters and seconds understood, we obtain ( , ) 5.0sin (0.10 4.0 ). y x t x t =- (g) Since / / , v k = = the tension is 2 1 2 2 2 1 2 (4.0g /cm)(4.0s ) 6400g cm/s 0.064 N....
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- Spring '07