The maximum velocity of a particle,executing simple harmonic motion with an amplitude $7 \ mm$,is $4.4 \ m/s$. The period of oscillation is .... $sec$

$A$ particle executes $S.H.M.$ with amplitude $a$ and time period $T$. The displacement of the particle when its speed is half of its maximum speed is $\frac{\sqrt{x} a}{2}$. The value of $x$ is $\ldots \ldots \ldots$

For a simple harmonic motion with amplitude $A$ and time period $T$,what is the velocity at $x = \frac{A}{2}$?

$A$ particle executes linear $S.H.M.$ with amplitude $A = 4 \text{ cm}$. The magnitude of velocity and acceleration is equal when it is at a distance $x = 3 \text{ cm}$ from the mean position. The time period of the oscillation is:

$A$ body is performing simple harmonic motion with amplitude $A$ and time period $T$. The variation of its acceleration $(f)$ with time $(t)$ is shown in the figure. If at time $t$,the velocity of the body is $v$,which of the following graphs correctly represents the variation of velocity $(v)$ with time $(t)$?

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