# A rope of mass m and length L is hanging from the ceiling.

• Natchanon
So is that what he means?In summary, to find the fundamental frequency in part C, you need to use the equation T = 4t_up, where t_up is the time it takes for the wave pulse to travel from the bottom to the top of the rope. This is because the period of the wave is related to the wavelength, which in this case is 4 times the length of the rope. This is due to the fact that the arc formed by the rope at its maximum deviation is 1/4 of the wavelength. This concept is related to constructive interference, where waves with the same phase can combine to form a larger wave.
Natchanon

## Homework Statement

Part C: Find fundamental frequency.

## Homework Equations

Tension(y) = μgy
v(y) = sqrt(gy)
Time it takes to travel from bottom to top = t_up = 2srqt(L/g)

## The Attempt at a Solution

I found part A and B, which are tension and velocity. I don't know how to find part C because velocity isn't constant so I can't just use f = (1/2L)sqrt(gy). I thought period was 2t_up, but that's wrong. The period is 4t_up, which I don't understand why. When the wave pulse reaches the ceiling, it changes phase and bounces back down, then it bounces up again with the same phase, then it changes phase to the same phase as the original pulse. Is that why we say period is 4t_up? When it comes back to the starting position with the same phase as the first incident wave?[/B]

Natchanon said:
The period is 4t_up, which I don't understand why.
I see this is marked as solved. In the absence of an explanation for that I assume it is a mistake.

the difficulty is in relating, one the one hand, the traveling of a wave pulse from one end to the other to, on the other hand, an initial and final configuration of the rope.
In the simplest oscillation mode, the rope will be at one time straight and vertical, at another, forming an arc with no inflection points from its attachment at the ceiling to the lower end, that lower end being at its maximum deviation. How do you think those two states relate to t_up?

haruspex said:
I see this is marked as solved. In the absence of an explanation for that I assume it is a mistake.

the difficulty is in relating, one the one hand, the traveling of a wave pulse from one end to the other to, on the other hand, an initial and final configuration of the rope.
In the simplest oscillation mode, the rope will be at one time straight and vertical, at another, forming an arc with no inflection points from its attachment at the ceiling to the lower end, that lower end being at its maximum deviation. How do you think those two states relate to t_up?
Because it is a free end, arc is 1/4 of the wavelength? L = λ/4 ==> λ = 4L.

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Natchanon said:
Because it is a free end, arc is 1/4 of the wavelength? L = λ/4 ==> λ = 4L.
That is certainly true. But is it obvious how that relates to the wave speed?

haruspex said:
That is certainly true. But is it obvious how that relates to the wave speed?
v = sqrt(gy) = 4L / T ==> T = 4L / sqrt(gy). But v isn't constant because it depends on y. Can I say that it takes 1 t_up to form λ/4, so it must take 4t_up to form λ, which means period = 4t_up?

Natchanon said:
v = sqrt(gy) = 4L / T ==> T = 4L / sqrt(gy). But v isn't constant because it depends on y. Can I say that it takes 1 t_up to form λ/4, so it must take 4t_up to form λ, which means period = 4t_up?
Then I'll use period = 4t_up to find the fundamental frequency.

Natchanon said:
Can I say that it takes 1 t_up to form λ/4
I think so, I was just saying it is not quite obvious.

My instructor said "The key concept here is that resonances if fundamentally a phenomenon of constructive interference. Think the guy in the kiddie pool video we saw in class, building a big wave by sending out waves that bounced off the edge of the pool and came back to the center, and then building by sending a new wave each time." I'm not sure how this applies here. I know that if we send a wave with the same phase as the reflected wave, we'll have one huge amplitude wave covering the entire length of the rope.

## Q: What is the formula for calculating the potential energy of the rope?

The potential energy of the rope can be calculated using the formula PE = mgh, where m is the mass of the rope, g is the acceleration due to gravity, and h is the height of the rope from the ground.

## Q: Does the length of the rope affect its potential energy?

Yes, the length of the rope does affect its potential energy. The longer the rope, the higher its potential energy will be due to the increase in height.

## Q: How can we determine the tension in the rope?

The tension in the rope can be determined by using the formula T = mg + ma, where m is the mass of the rope, g is the acceleration due to gravity, and a is the acceleration of the rope (which is equal to 0 if the rope is hanging still).

## Q: What happens to the potential energy of the rope if its mass is doubled?

If the mass of the rope is doubled, its potential energy will also double. This is because potential energy is directly proportional to mass, so an increase in mass will result in an increase in potential energy.

## Q: Is the potential energy of the rope affected by the material it is made of?

No, the potential energy of the rope is not affected by the material it is made of. It is only affected by its mass, height, and the acceleration due to gravity.

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