A wave pulse travels down a slinky. The mass of the slinky is m = 0.93 kg and is
ID: 1302833 • Letter: A
Question
A wave pulse travels down a slinky. The mass of the slinky is m = 0.93 kg and is initially stretched to a length L = 7.1 m. The wave pulse has an amplitude of A = 0.22 m and takes t = 0.406 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.44 Hz.
1)
What is the speed of the wave pulse?
m/s
2)
What is the tension in the slinky?
N
3)
What is the average speed of a piece of the slinky as a complete wave pulse passes?
m/s
4)
What is the wavelength of the wave pulse?
m
5)
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
Explanation / Answer
We have:
L = 7.1m
m = 0.93kg
f = 0.44Hz
1) v = distance/time = 7.1 / 0.406 = 17.48 m/s
2)T = m*v2 / L = 0.93 * (17.48)2 / 7.1 = 40.022
3) 4*amplitude*frequency = 4 x 0.22 x 0.44 = 0.387
4) wavelength = velocity*period (while period is 1/0.44Hz) = 17.48 x 2.272 = 39.72
5) double the distance new velocity = 2L / 0.406 = 34.97
Tension = (mass * (34.97)2) / 2L = 0.93 * (34.97)2/ 2 * 7.1 = 80.091
6) m / 2L = 0.93 / 2* 7.1 = 0.0654
7) t = distance over velocity = 2L / new velocity (34.97m/s) = 0.4060
8) wave length = period * new velocity = (1/0.44Hz) * (34.97m/s) = 79.477
9) the energy of the wave pulse depend on both frequency and amplitude.