$A$ spring executes $S.H.M.$ with mass $10 \ kg$ attached to it. The force constant of the spring is $10 \ N/m$. If at any instant its velocity is $40 \ cm/s$,the displacement at that instant is (Amplitude of $S.H.M.$ $= 0.5 \ m$) (in $m$)

$A$ particle is performing simple harmonic motion. Which of the following statements are correct?
$(i)$ Its velocity-displacement graph is parabolic in nature.
$(ii)$ Its velocity-time graph is sinusoidal in nature.
$(iii)$ Its velocity-acceleration graph is elliptical in nature.

$A$ simple harmonic oscillation is represented by $x = A \cos \left(\omega t + \frac{\pi}{4}\right)$. Its speed is maximum when $t$ equals

$A$ particle of mass $2 \, kg$ executing $SHM$ has an amplitude of $20 \, cm$ and a time period of $1 \, s$. Its maximum speed is ......... $m/s$.

$A$ particle performs linear $S.H.M.$ When the displacement of the particle from the mean position is $3 \ cm$ and $4 \ cm$,the corresponding velocities are $8 \ cm/s$ and $6 \ cm/s$ respectively. Its periodic time is

$A$ 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 . . . . . . $cm/s$.