The differential equation of a simple harmonic motion is given by $\frac{d^2y}{dt^2} + ky = 0$. What is the time period?

The displacement equation of a particle is $x = 3\sin 2t + 4\cos 2t$. The amplitude and maximum velocity will be respectively:

Study of which motion is required to understand many physical phenomena?

$A$ 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 $45^{\circ}$.

If $x=5 \sin \left(\pi t+\frac{\pi}{3}\right) \text{ m}$ represents the motion of a particle executing simple harmonic motion,the amplitude and time period of motion,respectively,are

$A$ particle is executing simple harmonic motion with an amplitude of $2 \,m$. The difference in the magnitudes of its maximum acceleration and maximum velocity is $4$. The time-period of its oscillation and its velocity when it is $1 \,m$ away from the mean position are respectively: