The displacement of a particle executing $SHM$ is given by $y = 5 \sin \left(4t + \frac{\pi}{3}\right)$. If $T$ is the time period and the mass of the particle is $2 \text{ g}$, the kinetic energy of the particle when $t = \frac{T}{4}$ is given by (in $\text{ J}$)

For the minimum time period condition,find the average potential energy between $t = 0$ to $t = 0.05 \ s$ (take $g = 10 \ m/s^2$).

$A$ 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 . . . . . . .

The potential energy function for a particle executing linear simple harmonic motion is given by $V(x) = k x^{2} / 2$,where $k$ is the force constant of the oscillator. For $k = 0.5 \; N m^{-1}$,the graph of $V(x)$ versus $x$ is shown in the figure. Show that a particle of total energy $1 \; J$ moving under this potential must 'turn back' when it reaches $x = \pm 2 \; m$.

$A$ linear harmonic oscillator of force constant $6 \times 10^5 \, N/m$ and amplitude $4 \, cm$ has a total energy of $600 \, J$. Select the correct statement.

For a body performing simple harmonic motion,its potential energy is $E_x$ at displacement $x$ and $E_y$ at displacement $y$ from the mean position. The potential energy $E_0$ at displacement $(x+y)$ is