$A$ particle performs $S.H.M.$ from the mean position. Its amplitude is $A$ and total energy is $E$. At a particular instant,its kinetic energy is $\frac{3E}{4}$. The displacement of the particle at that instant is:

The displacements of two particles of same mass executing $SHM$ are represented by the equations $x_1=4 \sin \left(10 t+\frac{\pi}{6}\right)$ and $x_2=5 \cos (\omega t)$. The value of $\omega$ for which the energies of both the particles remain same is (in $\text{ unit}$)

The kinetic energy of a particle executing simple harmonic motion at a displacement of $3 \ cm$ from the mean position is $4 \ mJ$. If the amplitude of the particle is $5 \ cm$,then the maximum force acting on the particle is (in $N$)

The total energy of a particle executing $SHM$ at an amplitude of $4 \, cm$ is $20 \, J$. What will be its total energy at a displacement of $x = 2 \, cm$ (in $J$)?

The total energy of a particle executing $S.H.M.$ is proportional to

$A$ linear harmonic oscillator of force constant $2 \times 10^6\, N/m$ and amplitude $0.01\, m$ has a total mechanical energy of $160\, J$. Which of the following statements is correct?