$A$ 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

Which of the following curves represents correctly the oscillation given by $y = y_0 \sin(\omega t - \phi)$,where $0 < \phi < 90^\circ$?

$A$ 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

Vertical displacement of a plank with a body of mass $m$ on it is varying according to the law $y = \sin \omega t + \cos \omega t$. The minimum value of $\omega$ for which the mass just breaks off the plank and the moment it occurs first after $t = 0$ are given by: ($y$ is positive vertically upwards)

The displacement of a simple harmonic oscillator after $3 \; s$ starting from its mean position is equal to half of its amplitude. The time period of the harmonic motion is $\dots \; s$.

$A$ particle performs simple harmonic motion between $x = -A$ and $x = +A$. How much time does it take to travel from $x = A$ to $x = A/2$?