An object of mass $3 \,kg$ is executing simple harmonic motion with an amplitude $\frac{2}{\pi} \,m$. If the kinetic energy of the object when it crosses the mean position is $6 \,J$, the time period of oscillation of the object is (in $\,s$)

An object of mass $0.2 \ kg$ executes simple harmonic motion along the $X-$ axis with a frequency of $\frac{25}{\pi} \ Hz$. At the position $x = 0.04 \ m$,the object has a kinetic energy of $0.5 \ J$ and a potential energy of $0.4 \ J$. The amplitude of oscillation in meters is equal to:

In a simple harmonic oscillator,at the mean position:

For any $S.H.M.$,the amplitude is $6\, cm$. If the instantaneous potential energy is half the total energy,then the distance of the particle from its mean position is .... $cm$.

$A$ particle starts executing simple harmonic motion $(SHM)$ of amplitude $a$ and total energy $E$. At any instant,its kinetic energy is $\frac{3E}{4}$. Then its displacement $y$ is given by:

For the minimum time period condition,find the average potential energy between $t = 0$ to $t = 0.05 \ s$ (take $g = 10 \ m/s^2$).