$A$ particle is executing $SHM$ with an amplitude $A$. The distance travelled by the particle in one time period is

For a particle performing linear $S.H.M.$,its average speed over one oscillation is ($A=$ amplitude of $S.H.M.$,$n=$ frequency of oscillation) (in $nA$)

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.

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

The motion of the particle is given by the equation $x = A \sin \omega t + B \cos \omega t$. The motion of the particle is:

Define simple harmonic motion $(SHM)$.