A wave pulse travels down a slinky. The mass of the slinky is m = 0.89 kg and is
ID: 1300677 • Letter: A
Question
A wave pulse travels down a slinky. The mass of the slinky is m = 0.89 kg and is initially stretched to a length L = 7.1 m. The wave pulse has an amplitude of A = 0.2 m and takes t = 0.426 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.47 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 Hookes Law)
N
6)
What is the new mass density of the slinky?
kg/m
7)
What is the new time it takes for a wave pulse to travel down the slinky?
s
8)
If the new wave pulse has the same frequency, what is the new wavelength?
m
9)
What does the energy of the wave pulse depend on?
the frequency
the amplitude
both the frequency and the amplitude
Explanation / Answer
The average transverse or longitudinal wave velocity over a complete cycle is zero.
Starting from the equilibrium point .. the first 1/2 cycle average velocity is zero ..
being two 1/4 cycle journeys in opposite directions (to travel back to the equilibrium position for the first time).
The magnitude of the velocity is the same (v) over each 1/4 cycle .. but in opposite directions each successive 1/4 .. +v + -v = 0 for each