Simple Harmonic Motion and Waves 1: 0, 12A 2: no 3: 14.2 s 4: 2 cycles, 4.39 J, 13.2 J, 4.39 m/s2x 5: 22.5 s, 29 J, x(t) = -2.41 cos(2.24t) m 6: 1.55 m 7: 2.46 hours 8: 675 km 9: yes 10: 400 m 11: 2.72 m 12: V2 = V201, 12 = V2x1 13: 12.5 N, 50 Hz, 350 Hz, yes 14: 450 Hz 15: 2.82 x 10 m, 2.82 m, 1.1 x 10 m, 1.1 x 10 m 16: 100 m11: You have a string with a tension of 5 N, a mass of 0.3 kg and a length of 4 in. If you oscillate one end of the string three times per second. what wavelength will the waves traveling down the string have? 12a: Two identical strings (same mass and length) are hung from the ceiling. String 1 supports a mass 1' while string 2 supports a mass 2:11. The masses are shifted back and forth horizontally which results in transverse waves propagating up the strings. Assume that M' ) liming. On which string is the wave propagation speed larger? By what factor? Explain. b: Assume that the masses are shifted back and forth at the same frequency on each string. ()11 which string is the wavelength longer? By what factor? Explain. 13a: A string on a guitar is oscillating at 200 Hz as shown (previous page). The equilibrium length of the string is 0.5 m and its mass is 0.0025 kg. Determine the tension in the string. b: Determine the fundamental frequency of this string. c: \\Vhat would be the frequency if the string were oscillating with six nodes (not counting the node at each end of the string]? Use the tension from part a. d: The string tension is changed to 50 X and the frequency is still 200 Hz. Is there a standing wave on the string new? If the answer is yes' then draw iL If the answer is no. then explain why' not. e: The string is currently at the position labeled A in the diagram. Assume that that position represents maximum displacement for the string. Draw and explain the appearance of the string exactly onefourth of an oscillation cycle later. 14: A guitar string of length L has a fundamental frequency of 150 Hz. The person playing the guitar places their nger at 1' : L/il so that there must be a node there now. He then plucks the string between 1: : 0 and J: : L/R. What is the new fundamental frequency