A 100 g block is connected to a horizontal massless spring of force constant `25.6(N)/(m)` As shown in Fig. the block is free to oscillate on a horizontal frictionless surface. The block is displaced 3 cm from the equilibrium position and , at `t=0`, it is released from rest at `x=0` It executes simple harmonic motion with the postive x-direction indecated in Fig. The position time `(x-t)` graph of motion of the block is as shown in Fig.
Position of the block as a function of time can now be expressed as

Options

A. `x=3cos(16t+(pi)/(2))cm`
B. `x=3cos(16t+(pi)/(3))cm`
C. `x=3.5cos(16t+(pi)/(6))cm`
D. `x=3.2cos(16t+(pi)/(4))cm`

Answer

C

Explanation

`omega=sqrt((k)/(m))=sqrt((25600)/(100))=16(rad)/(s)`
Using energy conservation `(1)/(2)kx^2+(1)/(2)mv^2=(1)/(2)kA'^(2)`
`A=sqrt(x^2+(m)/(k)v^2)=sqrt(9+3)=2sqrt3`
`x=Acos(omegat+phi)=2sqrt3cos(16t+phi)`
From boundary condition `t=0` `phi` can be obtained which is `(pi)/(6)`.
`x=2sqrt3cos(16t+(pi)/(6))cm`