The amplitude of an oscillating particle is $A$. When the velocity of the particle is one-third of its maximum velocity,determine the position of the particle.

The angular velocities of three bodies in simple harmonic motion are ${\omega _1}, {\omega _2}, {\omega _3}$ with their respective amplitudes as ${A_1}, {A_2}, {A_3}$. If all the three bodies have the same maximum velocity,then:

$A$ particle executes $S.H.M.$ of period $\frac{2 \pi}{\sqrt{3}} \text{ s}$ along a straight line $4 \text{ cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is (in $\text{ cm}$)

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 $rad/s$)

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

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