In $SHM$,the restoring force is $F = -kx$,where $k$ is the force constant,$x$ is the displacement,and $A$ is the amplitude of motion. Then,the total energy depends upon:

$A$ body is performing simple harmonic motion with an amplitude of $10 \, cm$. The velocity of the body is tripled by an air jet when it is at $5 \, cm$ from its mean position. The new amplitude of vibration is $\sqrt{x} \, cm$. The value of $x$ is . . . . . . .

For a particle executing $S.H.M.$,the potential energy $V$ versus displacement $x$ graph is given. The value of the restoring force constant is ..... $N/m$.

$A$ particle starts from the mean position and performs $S.H.M.$ with a period of $6 \ s$. At what time is its kinetic energy $50 \%$ of its total energy (in $s$)? $\left(\cos 45^{\circ} = \frac{1}{\sqrt{2}}\right)$

$A$ 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 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 . . . . . . .