If $x = a \sin \left( \omega t + \frac{\pi}{6} \right)$ and $x' = a \cos \omega t$,then what is the phase difference between the two waves?

$A$ particle executing a simple harmonic motion has a period of $6 \,s$. The time taken by the particle to move from the mean position to half the amplitude, starting from the mean position is

The phase of a particle executing simple harmonic motion is $\frac{\pi}{2}$ when it has:

$A$ particle is executing simple harmonic motion. Its amplitude is $A$ and time period is $5 \text{ sec}$. The time required by it to move from $x = A$ to $x = A/\sqrt{2}$ is . . . . . . sec.

$A$ particle performs $SHM$ about $x = 0$ such that at $t = 0$ it is at $x = 0$ and moving towards the positive extreme. The time taken by it to go from $x = 0$ to $x = \frac{A}{2}$ is ..... times the time taken to go from $x = \frac{A}{2}$ to $A$. The most suitable option for the blank space is

$A$ particle in $S.H.M.$ is described by the displacement function $x(t) = a\cos (\omega t + \theta )$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$. The angular frequency of the particle is $\pi \, rad/s$,then its amplitude is