If the only force acting on the rock is its weight, what is the magnitude of the rate of change of its angular momentum at this instant?

 The rate of chage of angular momentum = the net torque on the object. 
The net torque is the weight "mg" times the perpendicular distance from the line of action of the weight to the origin, x 
dL/dt = mgx = mgrCos(36.8) 

Or it is the component of the weight perpendicular to the position vector, mgCos(36.8) , times the magnitude of the position vector, r ; 
dL/dt = {mgCos(36.8}]r 

Obviously the same eq. 
Everything is given, so just plug in the values. The particle velocity is not needed to find rate of change of L.

Wavelength divides in half and speed stays the same.