A particle is executing simple harmonic motion with a time period of 3 s . At a position where the displacement of the particle is $60 \%$ of its amplitude. The ratio of the kinetic and potential energies of the particle is

The displacement of a particle executing simple harmonic motion is $y=A \sin (2 t+\phi) \mathrm{m}$, where $t$ is time in second and $\phi$ is phase angle. At time $t=0$, the displacement and velocity of the particle are 2 m and $4 \mathrm{~ms}^{-1}$. The phase angle, $\phi=$

The displacement of a damped oscillator is $x(t)=\exp (-0.2 t) \cos (3.2 t+\phi)$, where $t$ is time in second The time requirement for the amplitude of the oscillator to become $\frac{1}{e^{1.2}}$ times its initial amplitude is

Time period of a simple pendulum in air is $T$. If the pendulum is in water and executes SHM. Its time period is $t$. The value of $\frac{T}{t}$ is. (density of bob is $\frac{5000}{3} \mathrm{~kg} \mathrm{~m}^{-3}$ )