It is defined as the distance of the particle from the mean position at that instant.

Displacement in SHM at time ' t' is given by `x = A sin (omegat + phi)`.

As `sin (omegat + phi)` can take values between `-1` and `+1`, the displacement x can take values between `- A` and `+A`.

Suppose the position ofthe particle at `t=0` is `x_0` and its velocity is `v_0`. Thus at `t = 0`, `x = x_0` and `v = v_0`.

The acceleration of the particle at any instant is

`a=F/m=-(kx)/m=-omega^2x`

where `omega=sqrt(k/m)`

Thus, `(dv)/(dt)=-omega^2x....(1)`

`(dv)/(dx)\(dx)/(dt)=-omega^2x`

`vdv=-omega^2x`

The velocity of the particle is `v_0` when the particle is at `x = x_0`. It becomes `v` when the displacement becomes `x`. We can integrate the above equation and write

`int_(v_0)^v vdv=int_(x_0)^x -omega^2xdx`

`[(v^2)/2]_(v_0)^v=-omega^2[(x^2)/2]_(x_0)^x`

`v^2-v_0^2=-omega^2(x^2-x_0^2)`

`v=sqrt((v_0^2+omega^2x_0^2)-omega^2x^2)`

`v=omegasqrt(((v_0^2)/omega^2+x_0^2)-x^2)`

Here, `((v_0^2)/omega^2+x_0^2)=A`

the above equation becomes

`v=omegasqrt(A^2-x^2)`

We can write this equation as

`(dx)/(dt)=omegasqrt(A^2-x^2)`

`(dx)/(sqrt(A^2-x^2))=omegadt`

At time `t = 0` the displacement is `x = x_0` and at time `t` the displacement becomes `x`. The above equation can be integrated as

`int_(x_0)^x(dx)/(sqrt(A^2-x^2))=int_0^t omega dt`

`[sin^(-1)(x/A)]_(x_0)^x=[omegat]_0^t`

`sin^(-1)(x/A) - sin^(-1) ((x_0)/A) = omegat`

Here, `sin^(-1) ((x_0)/A)=phi`

`sin^(-1)(x/A) =omegat+phi`

`x=Asin(omegat+phi)....(2)`