Trouble with this integral and only an hour to solve it

[Brianjw]

Well first here is the question:

A deep-sea diver is suspended beneath the surface of Loch Ness by a cable of length h that is attached to a boat on the surface . The diver and his suit have a total mass of m and a volume of V. The cable has a diameter of d and a linear mass density of mu. The diver thinks he sees something moving in the murky depths and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions in the boat.

Calculate the tension in the cable a distance above the diver. The buoyant force on the cable must be included in your calculation. Take the free fall acceleration to be g.

I solve this one, I got:

$$F(x) = (\mu*x+m)*g-\rho_{water}*g*((d/2)^2*pi*x+V)$$

For the next one I can't solve it, the integral is nuts for me, I must be overlooking something

The speed of transverse waves on the cable is given by v = sqrt(F/mu). The speed therefore varies along the cable, since the tension is not constant. (This expression neglects the damping force that the water exerts on the moving cable.) Integrate to find the time required for the first signal to reach the surface. Take the free fall acceleration to be g.

From what I understand I have to plug in my F(x) from part b and integrate, but I'm not sure how to integrate it since its just so huge. Any ideas?

Thanks!

Brian

 

[ehild]

Well first here is the question:



$$F(x) = (\mu*x+m)*g-\rho_{water}*g*((d/2)^2*pi*x+V)$$

For the next one I can't solve it, the integral is nuts for me, I must be overlooking something

The speed of transverse waves on the cable is given by v = sqrt(F/mu). The speed therefore varies along the cable, since the tension is not constant. (This expression neglects the damping force that the water exerts on the moving cable.) Integrate to find the time required for the first signal to reach the surface. Take the free fall acceleration to be g.

From what I understand I have to plug in my F(x) from part b and integrate, but I'm not sure how to integrate it since its just so huge. Any ideas?

Thanks!

Brian

Write $$F/\mu$$ in the form $$F/\mu=ax+b$$ .

Remember that

$$v=x'=\frac{dx}{dt}$$ .

So you have got the differential equation

$$\frac{dx}{dt}=\sqrt{ax+b}$$

with the starting condition x=0 at t=0.
You can integrate it by separating the variables:

$$\int_0^h\frac{dx}{\sqrt{ax+b}}=\int_0^T{dt}=T$$

where T is the time needed for the signal to reach the surface and h is the length of the cable. The integrand at the left side is

$$(ax+b)^{-1/2}$$ .

I think you can do it...

ehild

 

[M98Ranger]

Hi Brian,

I understand that you are struggling with the second part of this problem and only have an hour to solve it. I can imagine the pressure you must be feeling, but don't worry, I'm here to help.

First of all, great job on solving the first part! It looks like you have a good understanding of the problem and have found the correct expression for the tension in the cable.

For the second part, you are correct that you need to integrate your expression for F(x) over the length of the cable to find the time it takes for the signal to reach the surface. I can see why this integral may seem daunting, but don't worry, there are ways to make it more manageable.

One approach you can take is to break up the integral into smaller, more manageable pieces. For example, you can split it into two integrals, one for the first term and one for the second term in your expression for F(x). You can also try using substitution or integration by parts to simplify the integral.

Another approach is to use numerical integration, where you use a computer program or calculator to approximate the integral. This can be a quicker and more efficient method, especially if you are short on time.

I recommend trying these different approaches and see which one works best for you. Don't forget to also include the buoyant force in your calculation. And remember, don't get too caught up in the details and take a step back if you feel stuck. Sometimes a fresh perspective can help.

Good luck and I hope you are able to solve the integral in time!