The displacement of a particle performing $S.H.M.$ is given by $Y = A \cos [\pi(t + \phi)]$. If at $t = 0$,the displacement is $y = 2 \text{ cm}$ and velocity is $v = 2\pi \text{ cm/s}$,the value of amplitude $A$ in $\text{cm}$ 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?

Why is the periodic time $T = \frac{2\pi}{\omega}$ taken for the function $f(t) = A \cos \omega t$?

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 displacement of a particle varies according to the relation $x = 3 \sin 100t + 8 \cos^2 50t$. Which of the following is incorrect about this motion?

The displacement of an oscillator is given by $x = a \sin \omega t + b \cos \omega t$,where $a, b$ and $\omega$ are constants. Then: