Select the wrong statement about simple harmonic motion $(S.H.M.)$.

$A$ particle moves with simple harmonic motion in a straight line. In the first $\tau \, s,$ after starting from rest,it travels a distance $a,$ and in the next $\tau \, s,$ it travels $2a$ in the same direction. Then:

$A$ particle of mass $m$ is under the influence of a force $F$ which varies with the displacement $x$ according to the relation $F = -kx + F_0$,where $k$ and $F_0$ are constants. The particle,when disturbed,will oscillate:

$A$ particle is executing simple harmonic motion with an amplitude $A$. The distance travelled by the particle in half time period is

The displacement of a particle is given at time $t$,by: $x = A \sin (-2 \omega t) + B \sin^2 \omega t$. Then,

The motion of a particle executing simple harmonic motion is described by the displacement function,$x(t) = A \cos (\omega t + \phi)$. If the initial $(t = 0)$ position of the particle is $1 \; cm$ and its initial velocity is $\omega \; cm/s$,what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \; s^{-1}$. If instead of the cosine function,we choose the sine function to describe the $SHM$: $x = B \sin (\omega t + \alpha)$,what are the amplitude and initial phase of the particle with the above initial conditions?