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

The maximum restoring force of a body executing $SHM$ is $\alpha$ and the total energy is $\beta$. Obtain its amplitude in terms of $\beta$ and $\alpha$.

$A$ body is performing $SHO$ with a total energy of $100\,J$. In the table below, column-$I$ shows the kinetic energy $(K)$ at a specific time, and column-$II$ shows the potential energy $(U)$ at that same time. Match them appropriately.

Column-$I$	Column-$II$
$(a)$ $K = 10\,J$	$(i)$ $U = 40\,J$
$(b)$ $K = 60\,J$	$(ii)$ $U = 90\,J$
	$(iii)$ $U = 50\,J$

An object of mass $0.5\, \text{kg}$ is executing simple harmonic motion. Its amplitude is $5\, \text{cm}$ and time period $T$ is $0.2\, \text{s}$. What will be the potential energy of the object at an instant $t = \frac{T}{4}\, \text{s}$ starting from the mean position? Assume that the initial phase of the oscillation is zero. (In $\text{J}$)

$A$ particle is executing simple harmonic motion starting from its mean position. If the time period of the particle is $1.5 \ s$,then the minimum time at which the ratio of the kinetic and total energies of the particle becomes $3: 4$ is

$A$ particle is executing $S.H.M.$ of amplitude $A$. When the potential energy of the particle is half of its maximum value during the oscillation,its displacement from the equilibrium position is