Maximum speed of a particle in simple harmonic motion is $v_{max}$. Then,the average speed of a particle in one complete oscillation is equal to:

$A$ particle is performing $S.H.M.$ with an amplitude $4 \,cm$. At the mean position, the velocity of the particle is $12 \,cm/s$. When the speed of the particle becomes $6 \,cm/s$, the distance of the particle from the mean position is:

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

$A$ body is executing simple harmonic motion with an angular frequency $2\,rad/s$. The velocity of the body at $20\,mm$ displacement,when the amplitude of motion is $60\,mm$,is ...... $mm/s$.

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 velocity of a particle executing simple harmonic motion along the $x$-axis is described by the equation $v^2 = 50 - x^2$, where $x$ represents displacement. If the time period of the motion is $\frac{x}{7} \ \text{s}$, the value of $x$ is . . . . . . .