"' a (8) A 102.3 g cord has an equilibrium length of 5.68 m. The cord is stretched horizontally to a length of 7.47 m, then vibrated at 28.2 Hz. This produces a standing wave with two antinodes. What is the spring constant of the cord? Submit Answer Tries 0/10 " 9: (9a) A 134.7 cm long, 8.27 g string, fixed at both ends, oscillates in its n = 3.00 mode with a frequency of 338.0 Hz and a maximum amplitude of 4.96 mm. What is the wavelength of the oscillation? |:| Submit Answer Tries 0/10 (9b) What is the tension in the string? |:| Submit Answer Tries 0/10 (8) The spring constant of the cord can be calculated using the formula for the speed of a wave on a string: v = JlT/u). where v is the speed of the wave. T is the tension in the string. and p is the linear density of the string (mass per unit length). The linear density of the cord is u = mlL. where m is the mass of the cord and L is its length. Substituting the given values. we get: u = 102.3 g / 5.68 m = 18.010g/m The distance between two antinodes in a standing wave with two antinodes is equal to half a wavelength, so: M2 = 7.47 m - 5.68 m =1.79 m )t=3.58m The frequency of the wave is l = 28.2 Hz. Using the formula for the speed of a wave on a string and substituting the known values. we get: v = {A = 28.2 Hz ' 3.58m :100.956m/s Finally. we can solve for the tension In the cord: T = uv2= 18.005 g/m ' (100.956m/s)2 = 183.55N/m Thereiore. the spring constant of the cord ls183.55N/m. (9a) The wavelength of the oscillation can be calculated using the formula for the wavelength of a standing wave on a string fixed at both ends: It = 2L. where L is the length of the string. n is the mode number. and A is the wavelength. Substituting the given values. we get: A = 2 ' 134.7 cm / 3.00 = 89.8 cm = 0.898 m Therefore, the wavelength of the oscillation is 0.898 in (9b) The tension in the string can be calculated using the formula for the frequency of a standing wave on a string fixed at both ends: f= i1/2Ll ' J(T/ui ' n. where T Is the tension in the string. [.1 is the linear density of the string. and n is the mode number. Solving for T. we get: T = i4L2' u '1'2l2 Substituting the given values. we get: T = (4 ~ (134.7 cm)2 ' 8.27 g/cm . (333.0 Hzl2) / (3.00) 2: 761.8KN Therefore. the tension in the string is 761.8KN