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

<br /> 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!

<br /> v=\sqrt{\frac{T}{\mu}}<br />

Therefore

<br /> \frac{dx}{dt}=\sqrt{\frac{T}{\mu}}<br />

So

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

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