The equation of a simple harmonic motion is $X = 0.34 \cos(3000t + 0.74)$,where $X$ and $t$ are in $mm$ and $sec$ respectively. The frequency of the motion is:

If the displacement of a particle is given by $x = 3 \sin (5\pi t) + 4 \cos (5\pi t) \text{ cm}$,what is the amplitude of the particle?

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

The displacement of a particle executing $S.H.M.$ is given by $x = 0.01 \sin 100 \pi(t + 0.05)$. The time period is ........ $s$.

The equation of a particle executing $SHM$ is given by $x = 3 \cos(\frac{\pi}{2}t)$ cm,where $t$ is in seconds. The distance travelled by the particle in the first $8.5$ seconds is:

The figure given below depicts two circular motions. The radius of the circle,the period of revolution,the initial position and the sense of revolution are indicated in the figures. Obtain the simple harmonic motions of the $x$-projection of the radius vector of the rotating particle $P$ in each case.