Velocity - time graph of a particle executing SHM is as shown in fig. Select the correct alternatives.A: at position 1, displacement of particle may be + ve or – veB: at position 2, displacement of particle is –veC: at position 3, acceleration of particle is +veD: at position 4, acceleration of particle is + ve

From the graph it is clear that at t = 0, the magnitude of velocity is maximum and negative and v-t gram is a negative cosine graph.i.e v=-v0cosωt=dxdtx=-v0ωsinωtSo at t = 0, the particle starts moving to the left from, equilibrium position o.So at 1, displacement is negative.At 2, velocity becomes zero for the first time, i.e the particle is at B and displacement is negative.At 3, velocity is positive ie. points to right, so the particle is somewhere between o and B and acceleration is directed towards 0 ie positive.At 4 the particle is some where between 0 and A and points towards A. So acceleration is directed towards 0 and negative. Hence option (2) is right.