A particle of mass 4 mg is executing simple harmonic motion along $X$-axis with an angular frequency of $40 \mathrm{rad} \mathrm{s}^{-1}$. If the potential energy of the particle is $V(x)=a+b x^2$, where $V(x)$ is in joule and $x$ is in metre, then the value of $b$ is

In a time $t$ amplitude of vibrations of a damped oscillator becomes half of its initial value, then the mechanical energy of the oscillator decreases by

For a particle executing simple harmonic motion, the kinetic energy of the particle at a distance of 4 cm from the mean position is $1 / 3$ rd of the maximum kinetic energy. The amplitude of the motion is