The velocity of a particle executing $SHM$ varies with displacement $(x)$ as $4v^2 = 50 - x^2$. The time period of oscillations is $\frac{x}{7} \ s$. The value of $x$ is $............$ (Take $\pi = \frac{22}{7}$)

The equations for the displacements of two particles in simple harmonic motion are $y_1=0.1 \sin \left(100 \pi t+\frac{\pi}{3}\right)$ and $y_2=0.1 \cos \pi t$ respectively. The phase difference between the velocities of the two particles at a time $t=0$ is

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 . . . . . . $m/s$.

The displacement of a body executing $SHM$ is given by $x = A \sin (2\pi t + \pi /3).$ The first time from $t = 0$ when the velocity is maximum is .... $\sec$

The velocity of a particle executing a simple harmonic motion is $13 \ m/s$,when its distance from the equilibrium position $(Q)$ is $3 \ m$ and its velocity is $12 \ m/s$,when it is $5 \ m$ away from $Q$. The frequency of the simple harmonic motion is

In an angular $SHM$,the angular amplitude of oscillation is $\pi \, rad$ and the time period is $0.4 \, s$. Calculate its angular velocity at an angular displacement of $\pi/2 \, rad$. (Result in $rad/s$)