$A$ particle of mass $250\,g$ executes a simple harmonic motion under a periodic force $F = (-25\,x)\,N$. The particle attains a maximum speed of $4\,m/s$ during its oscillation. The amplitude of the motion is $...........\,cm$.

$A$ $S.H.M.$ has amplitude $a$ and time period $T$. The maximum velocity will be

$A$ particle is performing simple harmonic motion with an amplitude of $0.06 \,m$ and a time period of $3.14 \,s$. The maximum velocity of the particle is . . . . . . $cm/s$.

The velocity-time diagram of a harmonic oscillator is shown in the figure. The frequency of oscillation is ..... $Hz$.

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

The velocity at the mean position of a particle executing $S.H.M.$ is $v$. What is the velocity of the particle at a distance equal to half of the amplitude?