The equation of motion of a particle of mass $1\,g$ is $\frac{d^2x}{dt^2} + \pi^2x = 0$,where $x$ is displacement (in $m$) from the mean position. The frequency of oscillation is (in $Hz$):

The displacement equation of a simple harmonic oscillator is given by $y = A \sin \omega t - B \cos \omega t$. The amplitude of the oscillator will be

$A$ 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 $45^o$.

What is the ratio between the distance travelled by the oscillator in one time period and the amplitude?

Fill in the blanks:
$1.$ The ratio of displacement at any position and ....... remains constant for a particle executing $SHM$.
$2.$ The radius of the reference circle is equal to the .......... of the oscillator.
$3.$ Increase in phase per second of $SHO =$ ......... .
$4.$ $SHO$ covers ......... distance in one periodic time.

The radius of the circle,the period of revolution,the initial position,and the sense of revolution are indicated in the figure. The $y$-projection of the radius vector of the rotating particle $P$ is: