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?

The equation of motion of a particle is $\frac{d^2y}{dt^2} + Ky = 0$,where $K$ is a positive constant. The time period of the motion is given by

The displacement equation of a simple harmonic oscillator is given by $y = A \sin \omega t - B \cos \omega t$. The amplitude of the oscillator will be

The figure shows the circular motion of a particle. The radius of the circle,the period,the sense of revolution,and the initial position are indicated in the figure. The simple harmonic motion of the $x$-projection of the radius vector of the rotating particle $P$ is

Define periodic motion and oscillatory motion.

$A$ point particle of mass $0.1 \ kg$ is executing $S.H.M.$ with an amplitude of $0.1 \ m$. When the particle passes through the mean position,its kinetic energy is $8 \times 10^{-3} \ J$. Obtain the equation of motion of this particle if the initial phase of oscillation is $45^{\circ}$.