The motion of a particle executing $S.H.M.$ is given by $x = 0.01 \sin 100\pi (t + 0.05)$,where $x$ is in metres and $t$ is in seconds. The time period is ..... $sec$.

If the function $\sin^2 \omega t$ (where $t$ is time in seconds) represents a periodic motion,then the period of the motion is:

The motion of a particle executing simple harmonic motion is described by the displacement function,$x(t) = A \cos (\omega t + \phi)$. If the initial $(t = 0)$ position of the particle is $1 \; cm$ and its initial velocity is $\omega \; cm/s$,what are its amplitude and initial phase angle? The angular frequency of the particle is $\pi \; s^{-1}$. If instead of the cosine function,we choose the sine function to describe the $SHM$: $x = B \sin (\omega t + \alpha)$,what are the amplitude and initial phase of the particle with the above initial conditions?

The figure shows the circular motion of a particle. The radius of the circle is $B$. The particle starts at $t=0$ from the positive $y$-axis and moves clockwise. The simple harmonic motion of the $x$-projection of the radius vector of the rotating particle is given by:

The figures correspond to two circular motions. The radius of the circle,the period of revolution,the initial position,and the sense of revolution (i.e.,clockwise or anti-clockwise) are indicated on each figure. Obtain the corresponding simple harmonic motions of the $x$-projection of the radius vector of the revolving particle $P$,in each case.

Given below are two statements. One is labelled as Assertion $(A)$ and the other is labelled as Reason $(R)$.
Assertion $(A) :$ Knowing initial position $x_0$ and initial momentum $p_0$ is enough to determine the position and momentum at any time $t$ for a simple harmonic motion with a given angular frequency $\omega$.
Reason $(R) :$ The amplitude and phase can be expressed in terms of $x_0$ and $p_0$. In the light of the above statements,choose the correct answer from the options given below $:$