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 –ve

B: at position 2, displacement of particle is –ve

C: at position 3, acceleration of particle is +ve

D: 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.

$\mathrm{v}=-{\mathrm{v}}_0\mathrm{cos\omega t}=\frac{\mathrm{dx}}{\mathrm{dt}}$

$\mathrm{x}=-\frac{{\mathrm{v}}_0}{\mathrm{\omega }}\mathrm{sin\omega t}$

So 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.