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

The displacement of a particle is given by the relation $x=4(\cos \pi t+\sin \pi t)$. The amplitude of the particle is

The displacement of a particle executing simple harmonic motion is given by $x=2 \cos (t)$ where $t$ is the time in seconds then the time period of the particle is