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Answer: added mass/moments of inertia is related to the amount of acceleration imparted on the surrounding fluid due to the particular mode of motion, considering the fluid to be ideal and inviscid. The surrounding fluid is `disturbed' by the `push' in the normal direction of the hull. If you take a semicircle, and try to rotate it about its centre, it will push no liquid in normal direction since the velocity at any point on the hull in its normal direction due to the imparted rotational motion is zero. So the added moment of inertia in roll for a semicircle is zero (taking the point of rotation at centre). Now, a typical ship section is basically not too far from a semicircle., which means, you can take a semicircle, and stretch it somewhat to get the ship section. Thus, a typical ship section, when you make it to rotate about its centre, the `push' or `normal velocity' of the hull at any point is relatively small. So the motion induced in the surrounding hull is also relatively small. This means, basically when you cause a ship hull to roll, the disturbance imparted on the surrounding water is relatively very small compared to say when you push it down (or impart a heave motion). This is the reason why roll added moment of inertia is relatively small for typical ship section. In fact, compared to rigid body roll inertia, it is only of the order of 5-10%. Since for motion analysis, added inertia always appears along with rigid body inertia, even if one neglects added inertia the error will be only marginal.
Answer: The question reads whether it is necessary to have uncoupled eqns. if mass and stiffness are symmetric. The question is not clear. Uncoupled eqns are simplified cases of coupled eqns, and results due to symmetry. Thus the question is not of necessity, perhaps the question is, if the mass and stiffness are symmetric, can one use uncoupled eqns. Well, besides just mass and stiffness, the hydrodynamic coefficients represented by added mass and damping also need to be symmetric. Since these quantities depend on external hull geometry, the question then reduces to what type of symmetry makes these hydro. coefficients symmetric. Basically centreplane symmetry causes the eqns. to get uncoupled from 6 modes to two sets of 3 modes. If there is further symmetry, then further reduction will happen. Thus if you take a hull which is symmetric about all the three planes through the origin, then the eqns. will reduce to single degree, provided mass matrix is symmetric (stiffness is dependent on external geometry - so it becomes symmetric automatically, or more appropriately all coupled stiffness coefficients reduce to zero). But the coupling may not be too strong in many cases, so often we can use a single degree uncoupled eqns. for a reasonably good estimate. To illustrate the coupling in physical terms, consider a ship being trimmed by rotating it about its CG. Since LCF is not necessarily about/below CG, this will automatically cause a parallel sinkage. In dynamic equivalence, thus, pitch will necessarily induce a little heave (and vice-versa). If the hull was fore-aft symmetric, then trim would not have caused heave, assuming of course mass matrix is symmetric.
Answer: It is true that for the two body coupled problem, there should be two peaks, each corresponding to the individual natural (resonant) frequency. Thus, in heave motion there will be one pronounced peak at its own natural frequency (where we have the heave resonant motion), and a second peak at pitch natural freq. (which happens because there pitch is largest, and due to coupling heave should technically attains a relatively large value). This is what happens in any case for a two-body mass-spring problem. But in practice, the coupling effect is not that strong between heave and pitch, and therefore mostly one does not see this in the response curve except for some very selective geometries and cases (even there the 2nd peak will be a relatively small one).