(i) In the system shown in figure, find the time period of vertical oscillations of the block A. Both the blocks A and B have equal mass of m and the force constant of the ideal spring is k. Pulley and threads are massless

(ii) In the arrangement shown in the figure the spring, string and the pulley are mass less. The force constant of the spring is k. A rope of mass per unit length equal to `lamda (kg m^(-1))` hangs from the string as shown. In equilibrium a length L of the rope is in air and its bottom part lies in a heap on the floor. The rope is very thin and size of the heap is negligible though the heap contains a fairly long length of the rope. The rope is raised by a very small distance and released. Show that motion will be simple harmonic and calculate the time period. Assume that the hanging part of the rope does not experience any force from the heap or the floor (For example there is no impact force while the rope hits the floor while moving downward and there is no impulsive pull when the vertical part jerks a small element of heap into motion).

Answer

(i) ` T=pi sqrt((5m)/(k)), (ii) T=2pi sqrt((lamda L)/(k+lmadag))`