$A$ simple pendulum performs simple harmonic motion about $X = 0$ with an amplitude $A$ and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be

Position of a $2 \,kg$ mass moving along the $X$-axis is given by $x=2 \cos (2 t) \,m$. Then the maximum kinetic energy of the mass in joules is

$A$ 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?

If a simple pendulum oscillates with an amplitude of $50\, mm$ and time period of $2\, s$,then its maximum velocity is .... $m/s$.

$A$ particle executing $SHM$ along a straight line has zero velocity at points $A$ and $B$ whose distances from $O$ on the same line $OAB$ are $a$ and $b$,respectively. If the velocity at the midpoint between $A$ and $B$ is $v$,then its time period is

Two identical pendulums oscillate with a constant phase difference $\frac{\pi}{4}$ and same amplitude. If the maximum velocity of one is $v$,the maximum velocity of the other will be ........