$A$ particle is performing simple harmonic motion with an amplitude of $4 \, cm$ and a time period of $12 \, s$. What is the ratio of the time taken by the particle to travel from its mean position to $2 \, cm$ to the time taken to travel from $2 \, cm$ to its extreme position?

$Y = A \sin (\omega t + \phi_{0})$ is the time-displacement equation of a $SHM$. At $t = 0$,the displacement of the particle is $Y = \frac{A}{2}$ and it is moving in the negative direction. Then the initial phase angle $\phi_{0}$ will be ...... .

$A$ particle performs simple harmonic motion with a period of $2 \ s$. The time taken by the particle to cover a displacement equal to half of its amplitude from the mean position is $\frac{1}{a} \ s$. The value of $a$ to the nearest integer is:

$A$ 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 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 ..... .

The displacement-time equation of a particle executing $SHM$ is $x = A \sin \left( \omega t + \frac{\pi}{6} \right)$. The time taken by the particle to go directly from $x = -\frac{A}{2}$ to $x = +\frac{A}{2}$ is