Simple Nature - Light and Matter

so the energy in the spring isU = 1 2 k ( Lb x ) 2= kL22b 2 x 2 ,and the k we have to put in T = 2π √ m/k isThe result isd 2 Udx 2 = kL2b 2 .√mb 2T = 2πkL 2= 2πb√mL kThe leverage of the lever makes it as if the spring was stronger,and decreases the period of the oscillations by a factor of b/L.Water in a U-shaped tube example 24⊲ What is the period of oscillation of the water in figure e?⊲ In example 13 on p. 89, we found U(y) = ρgAy 2 , so the “springconstant,” which really isn’t a spring constant here at all, isk = d2 Udy 2= 2ρgA .This is an interesting example, because k can be calculated withoutany approximations, but the kinetic energy requires an approximation,because we don’t know the details of the patternof flow of the water. It could be very complicated. There willbe a tendency for the water near the walls to flow more slowlydue to friction, and there may also be swirling, turbulent motion.However, if we make the approximation that all the water moveswith the same velocity as the surface, dy/dt, then the mass-ona-springanalysis applies. Letting L be the total length of the filledpart of the tube, the mass is ρLA, and we haveT = 2π √ m/k√ρLA= 2π2ρgA√L= 2π2g.e / Water in a U-shaped tube.This chapter is summarized on page 947. Notation and terminologyare tabulated on pages 941-942.Section 2.5 Oscillations 117