The equation of a particle executing simple harmonic motion is given by $x = \sin \pi (t + 1/3) \, m$. At $t = 1 \, s$,the speed of the particle will be .......... $cm \, s^{-1}$ (Given: $\pi = 3.14$)

$A$ 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 simple harmonic motions are represented by equations $y_1 = 4 \sin(10t + \phi)$ and $y_2 = 5 \cos(10t)$. What is the phase difference between their velocities?

An object is executing simple harmonic motion with an angular frequency $\omega$. If the maximum velocity is $v_{\max}$,then the maximum acceleration of the object is:

The velocity of a particle in simple harmonic motion at displacement $y$ from the mean position is

$A$ particle is performing simple harmonic motion with angular frequency $\omega$ and amplitude $A$. If $a$ is acceleration and $v$ is speed at any instant,then the graph showing the correct variation between $v^2$ and $x^2$ is (where $x$ is displacement):