$A$ particle executing $S.H.M.$ with an amplitude of $4 \, cm$ and a time period $T = 4 \, s$. The time taken by it to move from the positive extreme position to half the amplitude is ..... $s$.

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?

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)

Two bodies performing $S.H.M.$ have the same amplitude and frequency. Their positions and directions of motion at a certain instant are as shown in the figure. The phase difference between them is

The initial phase of a body executing $SHM$ is $\frac{\pi}{4}$. What will be its phase at the end of $10$ oscillations?

$A$ particle executes $SHM$ with amplitude $0.2 \,m$ and time period $24 \,s$. The time required for it to move from the mean position to a point $0.1 \,m$ from the mean position is (in $\,s$)