The displacement of a particle performing linear $S.H.M.$ is given by $y = A \cos[\pi(t + \phi)]$. If at $t = 0$, the displacement is $y = 2 \text{ cm}$ and velocity is $2\pi \text{ cm/s}$, the value of amplitude $A$ in $\text{cm}$ is

$A$ particle of mass $0.4 \text{ kg}$ executes simple harmonic motion of amplitude $0.4 \text{ m}$. When it passes through the mean position,its kinetic energy is $256 \times 10^{-3} \text{ J}$. If the initial phase of the oscillation is $\pi / 4$,then the equation of its motion is

$A$ particle executing $SHM$ has a maximum speed of $0.5 \ m s^{-1}$ and a maximum acceleration of $1.0 \ m s^{-2}$. The angular frequency of oscillation is

Why is the periodic time $T = \frac{2\pi}{\omega}$ taken for the function $f(t) = A \cos \omega t$?

$A$ particle executes simple harmonic motion with an amplitude '$A$'. The distance travelled by it in one periodic time is

What is the ratio between the distance travelled by the oscillator in one time period and the amplitude?