How Does a Buoy Respond to a 4m High Wave?

[iloveannaw]

there is buoy bobbing gently on the open sea. we need to find
1) average density
2) natural frequency
3) amplitude when the buoy is struck by a wave 4m high (peak to trough) with period 5sec.

I've worked out parts 1 and 2:

[tex]\rho_B = \rho_w \frac{h}{H}[/tex] (h = height submerged, H= total height)

[tex]\omega_0 = \sqrt{\frac{\rho_w F g}{m}} = \sqrt{\frac{g}{h}}[/tex] with F = cross-sectional areabut I don't know about the last part - I tried this:

[tex]x(t) = Acos(\omega_0 t) + 2cos(\frac{2\pi}{5}t)[/tex]

where A equals x from the differential equation like so:

[tex]\ddot{x} = \sqrt{\frac{g}{h}}x \Rightarrow A = x = \sqrt{\frac{h}{g}} \ddot{x} [/tex]

letting the circular functions equal 1 gives an amplitude of

[tex]x(t) = \sqrt{\frac{h}{g}} \ddot{x} + 2[/tex]

which doesn't seem right to me- please help!

 

[anathema]

For the last part, you can use the equation for the amplitude of a forced harmonic oscillator:

A = \frac{F_0}{m\sqrt{(\omega_0^2 - \omega^2)^2 + (\beta\omega)^2}}

where F_0 is the amplitude of the forcing function (in this case, the wave), m is the mass of the buoy, \omega_0 is the natural frequency, \omega is the angular frequency of the forcing function (in this case, \frac{2\pi}{5}), and \beta is the damping coefficient (which we can assume to be negligible for a buoy on the open sea).

Plugging in the values, we get:

A = \frac{mg}{m\sqrt{(\frac{g}{h} - (\frac{2\pi}{5})^2)^2 + 0}}

Simplifying, we get:

A = \frac{5g}{\sqrt{(g - (\frac{2\pi h}{5})^2)^2}}

For a wave with a height of 4m, the amplitude would be:

A = \frac{5g}{\sqrt{(g - (\frac{2\pi \cdot 4}{5})^2)^2}} \approx 3.96 m

So the amplitude of the buoy when struck by a wave 4m high would be approximately 3.96m.