The periodic time of a body executing simple harmonic motion is $3 \, s$. After how much interval from time $t = 0$,will its displacement be half of its amplitude?

$A$ simple harmonic oscillator has an amplitude $a$ and time period $T$. The time required by it to travel from $x = a$ to $x = a/2$ is

$A$ particle executes simple harmonic motion represented by the displacement function $x(t) = A \sin (\omega t + \phi)$. If the position and velocity of the particle at $t = 0 \, s$ are $2 \, cm$ and $2 \omega \, cm \, s^{-1}$ respectively,then its amplitude is $x \sqrt{2} \, cm$,where the value of $x$ is ..... .

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

The time taken by a particle executing simple harmonic motion of period $T$ to move from the mean position to half the maximum displacement 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)