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 is executing simple harmonic motion with time period $2 \ s$ and amplitude $1 \ cm$. If $D$ and $d$ are the total distance and displacement covered by the particle in $12.5 \ s$,then $\frac{D}{d}$ is:

The graphs in the figure show that a quantity $y$ varies with displacement $d$ in a system undergoing simple harmonic motion. Which graph best represents the relationship obtained when $y$ is the time?

The displacement $x$ (in metres) of a particle performing simple harmonic motion is related to time $t$ (in seconds) as $x = 0.05 \cos \left( 4 \pi t + \frac{\pi}{4} \right)$. The frequency of the motion will be ..... $Hz$.

$A$ 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

$A$ body oscillates with $SHM$ according to the equation (in $SI$ units):
$x = 5 \cos (2 \pi t + \pi / 4)$
At $t = 1.5 \, s$,calculate the:
$(a)$ displacement
$(b)$ speed
$(c)$ acceleration of the body.