$A$ body starting at $t=0$ from the origin oscillates simple harmonically with a period of $4 \ s$. After what time will its kinetic energy be $75 \%$ of its total energy?

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

The displacement of a particle, executing simple harmonic motion with time period $T$, is expressed as $x(t) = A \sin \omega t$, where $A$ is the amplitude. The maximum value of potential energy of this oscillator is found at $t = T / (2 \beta)$. The value of $\beta$ is . . . . . . .

$A$ body is executing simple harmonic motion. At a displacement $x$,its potential energy is $E_1$ and at a displacement $y$,its potential energy is $E_2$. The potential energy $E$ at a displacement $(x + y)$ is

The amplitude of a particle executing $SHM$ is $3 \text{ cm}$. The displacement at which its kinetic energy will be $25\%$ more than its potential energy is (in $\text{cm}$):

The potential energy of a simple harmonic oscillator of mass $2 \,kg$ at its mean position is $5 \,J$. If its total energy is $9 \,J$ and amplitude is $1 \,cm$, then its time period is