The maximum velocity of a particle performing simple harmonic motion is $6.28 \text{ cm s}^{-1}$. If the length of its path is $8 \text{ cm}$, then what is its period (in $\text{ s}$)?

In an angular $SHM$,the angular amplitude of oscillation is $\pi \, rad$ and the time period is $0.4 \, s$. Calculate its angular velocity at an angular displacement of $\pi/2 \, rad$. (Result in $rad/s$)

$A$ particle executing simple harmonic motion has a maximum speed of $40 \,ms^{-1}$ and maximum acceleration of $60 \,ms^{-2}$. The period of oscillation is

$A$ particle performs simple harmonic motion with a time period of $16 \ s$. At a time $t=2 \ s$, the particle passes through the origin and at $t=4 \ s$ its velocity is $4 \ m/s$. The amplitude of the motion is

$A$ simple harmonic oscillator has a period of $0.01 \, s$ and an amplitude of $0.2 \, m$. The magnitude of the velocity in $m \cdot s^{-1}$ at the centre of oscillation is:

$A$ particle is executing $S.H.M.$ and its velocity $v$ is related to its position $x$ as $v^2 + ax^2 = b$,where $a$ and $b$ are positive constants. The frequency of oscillation of the particle is ..........