$A$ particle executes simple harmonic motion along a straight line with an amplitude $A$. The potential energy is maximum when the displacement is

The acceleration $A$ and time $t$ of a body in $S.H.M.$ is given by the curve shown below. Then the corresponding graph between kinetic energy $(K.E.)$ and time $t$ is correctly represented by:

The amplitude of an $SHM$ particle is $4 \, cm$. At what distance from the mean position will the potential energy and kinetic energy be equal?

At a given point of time,the displacement of a simple harmonic oscillator is given by $y = A \cos(30^{\circ})$. If the amplitude is $40 \, cm$ and the kinetic energy at that time is $200 \, J$,the value of the force constant is $1.0 \times 10^{x} \, Nm^{-1}$. The value of $x$ is ......

$A$ particle is vibrating in $S.H.M.$ with an amplitude of $4 \,cm$. At what displacement from the equilibrium position is its energy half potential and half kinetic?

$A$ particle starting from mean position executes simple harmonic motion with a period $8 \ s$. The minimum time in which its potential energy becomes half of the total energy is . . . . . . . (in $s$)