There are two ways of seeing why the higher order modes have faster frequencies:
1: One way is that since each half of the object is flexing in the opposite direction of the other half, there are effectively two motions going on at once, each involving only half the mass of the object.
Therefore, since there is less mass to move, greater accelerations are imparted by the flexural stress, and the object will return to its undeformed shape sooner. Of course, it then continues to move into the opposite flexion and repeatedly return to the undeformed shape. Therefore, the frequency is higher when the mode shape involves several small curves in alternating directions.
2:The other way to see it is that for a given amplitude of displacement, the curve is sharper. For instance, consider a simple shape such as a rod of length L. Vibrating in one large bow shape, the equation for the deflection is y = A*sin(pi*x/L) where x is the distance from one end, y is the deflection at location x, and A is the maximum deflection. The force which is trying to return the rod to its original straight shape arises from the beam shear V, by V(x+dx) - V(x) = F, acting on a mass wdx, where w is the beam mass per unit length. This gives an acceleration F/m = (dV/dx)/w. This is of course M/w. Now consider a shape in which the rod bends into a full sine wave with the same amplitude, y = A*sin(2pi*x/L). Since the moment is proportional to the second derivative of the deflection, the moment here will be 4 times as great as in the original case (because the extra factor of 2 arises from each differentiation). Therefore, there is more force per unit mass, hence more acceleration, and again, the rod returns to its straight shape sooner.
