Solution of u´´´´ = freq² u with boundary condition u´´´(1)=u´´´(-1)= u´´(1)=u´´(-1)=0. Here freq is proportional to the frequency of oscillation.
Solution of Δ Δ u = freq² u with as boundary conditions that Δ u and its gradient vanish on the edge of the disc. Here Δ is the Laplacian.
Picture for square plate obtained from solutions for the thin bar by the method of Wheatstone.
Many more pictures.