A wave pulse travels down a slinky. The mass of the slinky is m = 0.92 kg and is
ID: 1501059 • Letter: A
Question
A wave pulse travels down a slinky. The mass of the slinky is m = 0.92 kg and is initially stretched to a length L = 7.3 m. The wave pulse has an amplitude of A = 0.27 m and takes t = 0.47 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.46 Hz. What is the tension in the slinky? What is the wavelength of the wave pulse? Now the slinky is stretched to twice its length (but the total mass does not change). What is the new tension in the slinky? (assume the slinky acts as a spring that obeys Hooke’s Law) If the new wave pulse has the same frequency, what is the new wavelength?
Explanation / Answer
1.) v = distance/time = 7.3 / .47 = 15.53m/s
T = m*v^2 / L = 0.92 (15.53)^2 /7.3 = 30.4 N
(2) wavelength = velocity /frequency =15.53/0.46= 33.8 m
(3)when we double the lenght, new velocity = 2L / .47 = 31.06
Tension = (mass * 31.06^2) / 2L = (0.92 * 31.06^2) / (2*7.3 )=60.79 N
(4)wavelength = velocity /frequency =30.06/0.46= 67.5 m