$A$ particle moves on the $x-$axis according to the equation $x = x_0 \sin^2 \omega t$. Is the motion simple harmonic?

Equations $y_1 = A \sin \omega t$ and $y_2 = \frac{A}{2} \sin \omega t + \frac{A}{2} \cos \omega t$ represent $S.H.M.$ The ratio of the amplitudes of the two motions is

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?

Define the amplitude of $SHM$.

$A$ simple harmonic motion is represented by $y = 5(\sin 3\pi t + \sqrt{3} \cos 3\pi t) \ cm$. The amplitude and time period of the motion are:

If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4v^2 = 25 - x^2$, then the time period is