What is the Time Required for a Diver's Signal to Reach the Surface?

[kudoushinichi88]

A deep-water diver is suspended beneath the water surface by a 100-m long cable. The diver and his suit have a total mass of 120-kg and a volume of 0.0800-m³. The cable has a diameter of 2.00cm and a linear mass density of $\mu=1.10\mbox{kg/m}$.

a) Calculate the tension in the vable a distance x above the diver

Tension in cable, T

$$T=m_{diver}g+m_{cable}g-\rho g V_{diver}-\rho g V_{cable}$$

Subbing in the values,

$$T=392.4+10.791x-0.981\pi x$$

b) The diver thinks he sees something approaching and jerks the end of the cable back and forth to send transverse waves up the cable as a signal to his companions. Calculate the time required for the first signal to reach the surface. Ignore the damping of the water.

Speed of the wave on the cable, v

$$v=\sqrt{\frac{T}{\mu}}$$

so the time taken to reach the surface t,

$$t=s\sqrt{\frac{\mu}{T}}$$

Subbing the values and the result from a), and integrating over distance x,

$$t=\int_{0}^{100}100\sqrt{\frac{1.1}{392.4+10.791x-0.981\pi x}}dx$$

and I finally get an answer of

$$t=389\mbox{s}$$

Is there anything wrong with my steps?

 

[kudoushinichi88]

uh... no.

Oh wait. hang on a second. I think I'm doing stuff without thinking. XD

Please hold on!

 

[kudoushinichi88]

Ahah!

$$v=\sqrt{\frac{T}{\mu}}$$

Therefore

$$\frac{dx}{dt}=\sqrt{\frac{T}{\mu}}$$

So

$$t=\int_{0}^{100}\sqrt{\frac{1.1}{392.4+10.791x-0.981\pi x}}dx$$

Which gives a value of... wow 3.89 seconds? That's fast?I still don't get your cookie metaphor though. =P