$A$ $S.H.M.$ is represented by $x = 5\sqrt{2} (\sin 2\pi t + \cos 2\pi t).$ The amplitude of the $S.H.M.$ is .... $cm$.

$A$ point particle of mass $0.1 \ kg$ is executing $S.H.M.$ with an amplitude of $0.1 \ m$. When the particle passes through the mean position,its kinetic energy is $8 \times 10^{-3} \ J$. Obtain the equation of motion of this particle if the initial phase of oscillation is $45^{\circ}$.

The displacement of a particle along the $x$-axis is given by $x = a \sin^2 \omega t$. The motion of the particle corresponds to:

$A$ particle moves such that its acceleration $a$ is given by $a = -bx$,where $x$ is the displacement from the equilibrium position and $b$ is a constant. The period of oscillation is

The differential equation of a simple harmonic motion is given by $\frac{d^2y}{dt^2} + ky = 0$. What is the time period?

If the displacement $(x)$ and velocity $(v)$ of a particle executing simple harmonic motion are related through the expression $4v^2 = 25 - x^2$,then the time period is