Solving Wave Pulse Problems in a Slinky: Speed & Tension

[ctwokay]

Homework Statement

A wave pulse travels down a slinky. The mass of the slinky is m = 0.86 kg and is initially stretched to a length L = 6.5 m. The wave pulse has an amplitude of A = 0.22 m and takes t = 0.494 s to travel down the stretched length of the slinky. The frequency of the wave pulse is f = 0.48 Hz.

1.What is the average speed of a piece of the slinky as a complete wave pulse passes?

2.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)

Homework Equations

v=λ*f
ω=2*∏*f
v=√(F/μ)

The Attempt at a Solution

1. Tried to use v=λ*f but can't seem to find k for λ

2. Tried to use v=√(F/μ) to combine with hooke's law F=-kx but could not find the answer.

I not sure whether I use the correct equation.

 

[anathema]

Can someone help me?

Thank you for your post. It seems like you are on the right track with your equations, but there are a few things to consider in order to find the solutions to these problems.

1. The average speed of a piece of the slinky can be found by dividing the total distance traveled by the total time it took to travel that distance. In this case, the total distance traveled is the length of the slinky, which is 6.5 m. The total time is given as 0.494 s. So, the average speed would be 6.5 m/0.494 s = 13.17 m/s.

2. To find the new tension in the slinky, we can use Hooke's Law, which states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed from its equilibrium position. In this case, the slinky is stretched to twice its length, so the new length would be 2L = 13 m. We also know the mass of the slinky (m = 0.86 kg) and the frequency (f = 0.48 Hz). Now, we can use the equation v=√(F/μ) to solve for the new tension. Remember that μ is the linear mass density of the slinky, which can be found by dividing the mass by the length (μ = m/L). So, the equation becomes v=√(F/(m/L)). We also know that v=λ*f, so we can substitute that into the equation to get λ*f=√(F/(m/L)). Rearranging the equation, we get F = (m/L)*(λ*f)^2. Now, we can use the given values of m, L, λ and f to find the new tension in the slinky.

I hope this helps. If you have any further questions, please feel free to ask. Good luck with your calculations!