The amplitude of a particle executing $SHM$ is made three-fourth keeping its time period constant. Its total energy will be

$A$ loaded vertical spring executes $S.H.M.$ with a time period of $4\; sec$. The difference between the kinetic energy and potential energy of this system varies with a period of ........ $sec$.

The total mechanical energy of a particle in $SHM$ is

Two independent harmonic oscillators of equal mass are oscillating about the origin with angular frequencies $\omega_1$ and $\omega_2$ and have total energies $E_1$ and $E_2$,respectively. The variations of their momenta $p$ with positions $x$ are shown in the figures. If $\frac{a}{b}= n^2$ and $\frac{a}{R}= n$,then the correct equation$(s)$ is(are):
$(A) E_1 \omega_1 = E_2 \omega_2$
$(B) \frac{\omega_2}{\omega_1} = n^2$
$(C) \omega_1 \omega_2 = n^2$
$(D) \frac{E_1}{\omega_1} = \frac{E_2}{\omega_2}$

$A$ particle of mass $4 \,kg$ is executing $SHM$. Its displacement is given by the equation $y=8 \cos [100 t+\pi / 4] \,cm$. Its maximum kinetic energy is (in $\,J$)

Obtain the expressions for kinetic energy,potential energy,and total energy in simple harmonic motion.