For a particle executing simple harmonic motion $(SHM)$,at its mean position

$A$ particle executing simple harmonic motion has a maximum acceleration of $24 \ m/s^2$ and a maximum velocity of $16 \ m/s$. What is its amplitude?

$A$ particle performs $S.H.M.$ with amplitude $A$. Its speed is tripled at the instant when it is at a distance of $\frac{2A}{3}$ from the mean position. The new amplitude of the motion is

$A$ particle performs simple harmonic oscillation of period $T$ and the equation of motion is given by $x = a \sin(\omega t + \pi/6)$. After the elapse of what fraction of the time period will the velocity of the particle be equal to half of its maximum velocity?

$A$ particle in $SHM$ is described by the displacement equation $x(t) = A\cos(\omega t + \theta)$. If the initial $(t = 0)$ position of the particle is $1 \, cm$ and its initial velocity is $\pi \, cm/s$,what is its amplitude? The angular frequency of the particle is $\pi \, s^{-1}$.

The displacement of a body executing $SHM$ is given by $x = A \sin (2\pi t + \pi /3).$ The first time from $t = 0$ when the velocity is maximum is .... $\sec$