The displacement of a particle executing linear $S.H.M.$ is given by $x = 0.25 \sin(11t + 0.5) \ m$. The period of $S.H.M.$ is (take $\pi = \frac{22}{7}$):

The motion of a particle represented by $y = \sin \omega t - \cos \omega t$ is

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

The displacement $x$ (in metres) of a particle performing simple harmonic motion is related to time $t$ (in seconds) as $x = 0.05 \cos \left( 4 \pi t + \frac{\pi}{4} \right)$. The frequency of the motion will be ..... $Hz$.

Two particles executing $SHM$ along a straight line have the same amplitude $A$ and time period $T$. At $t=0$,one particle is at a displacement $+A$ and another is at a displacement $-\frac{A}{2}$ and they are approaching towards each other. They cross each other after a time.

Define periodic time and frequency,their units,and the relationship between them.