$A$ simple pendulum performs simple harmonic motion about $x=0$ with an amplitude $a$ and time period $T$. The speed of the pendulum at $x=a/2$ will be

The displacement of a particle executing $SHM$ is given by $x = 10 \sin \left(\omega t + \frac{\pi}{3}\right) \ m$. The time period of motion is $3.14 \ s$. The velocity of the particle at $t = 0$ is . . . . . . $m/s$.

$A$ body is executing $S.H.M.$ When its displacement from the mean position is $4 \, cm$ and $5 \, cm$,the corresponding velocity of the body is $10 \, cm/sec$ and $8 \, cm/sec$. Then the time period of the body is

The displacement of a particle in $S.H.M.$ is given by $x = A \cos(\omega t + \pi/6)$. At what time will its speed be maximum?

$A$ particle performing $S.H.M.$ starts from the equilibrium position and its time period is $12 \ s$. After $2 \ s$,its velocity is $\pi \ m/s$. The amplitude of the oscillation is: $[\sin 30^{\circ}=\cos 60^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3}/2]$

The angular velocities of three bodies in simple harmonic motion are ${\omega _1}, {\omega _2}, {\omega _3}$ with their respective amplitudes as ${A_1}, {A_2}, {A_3}$. If all the three bodies have the same maximum velocity,then: