Sir, Please help me with this problem. 1) Two simple pendulums of length 1metre and 16 metre respectively are both given small displacements in the same direction at the same instant. They will again be in phase after the shorter pendulum has completed n oscillations. What is the value of n? I solved a similar kind of problem in the following way: 2) Two pendula of lengths 121cm and 100cm start vibrating. At some instant they are in mean position in the same phase. After how many vibrations of the shorter pendulum, the two will be in the same phase in the mean position? Let L(1) = 1.21 m T(1) = 2(pie) (L(1)/g)^1/2 [2 pie root L one by g) T(1) = 2.2 seconds L(2) = 1.00 m T(2) = 2(pie) (L(2)/g)^1/2 [Read as 2 pie root L two by g) T(2) = 2 seconds Since when they are in phase again the total time elapsed is the same, T(1) * X(1) = T(2) * X(2) X(2)/X(1) = T(1)/T(2) = 2.2/2 = 1.1 = 11/10 So they will again be in phase when the shorter pendulum has completed 11 oscillations. But this method doesn’t work with the first problem. Using the above method in the first problem, I get the answer as 4. But the answer given in my book is 4/3.