The displacement of particle in S.H.M. is $\mathrm{x}=\mathrm{A} \cos (\omega \mathrm{t}+\pi / 6)$. Its speed will be maximum at time $\left(\sin 90^{\circ}=1\right)$

An object of mass 0.2 kg executes simple harmonic oscillations along the x -axis with frequency of $\left(\frac{25}{\pi}\right) \mathrm{Hz}$. At the position $x=0.04 \mathrm{~m}$, the object has kinetic energy 1 J and potential energy 0.6 J . The amplitude of oscillation is

The motion of the particle is given by the equation $\mathrm{x}=\mathrm{A} \sin \omega \mathrm{t}+\mathrm{B} \cos \omega \mathrm{t}$.

The motion of the particle is

A particle is executing S.H.M. of amplitude ' $A$ '. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is