The time taken by a particle executing simple harmonic motion of period '$$\mathrm{T}$$', to move from the mean position to half the maximum displacement is

In a medium, the phase difference between two particles separated by a distance '$$x$$' is $$\left(\frac{\pi}{5}\right)^{\text {c }}$$. If the frequency of the oscillation of particles is $$25 \mathrm{~Hz}$$ and the velocity of propagation of the waves is $$75 \mathrm{~m} / \mathrm{s}$$, then the value of $$x$$ is

A particle starts oscillating simple harmonically from its mean position with time period '$$T$$'. At time $$t=\frac{T}{12}$$, the ratio of the potential energy to kinetic energy of the particle is $$\left(\sin 30^{\circ}=\cos 60^{\circ}=0.5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right)$$

The displacement of a particle performing S.H.M. is given by $$x=5 \sin (3 t+3)$$, where $$x$$ is in $$\mathrm{cm}$$ and $$t$$ is in second. The maximum acceleration of the particle will be