Two cars, A and B, start from rest at the same point. Car A moves with constant acceleration 'a' and car B moves with constant acceleration '2a'. After how much time will the distance between them be 'S'?

A simple pendulum performs simple harmonic motion about $\;X = 0$ with an amplitude A and time period $T$. The speed of the pendulum at $X = \frac{A}{2}$ will be

A particle's velocity is given by $v = 4 - 2t$, where $t$ is in seconds and $v$ is in m/s. If the particle starts at $x = 2$ m at $t = 0$, at what time does the particle instantaneously change its direction of motion?

A particle executes S.H.M. with a period of 6 second and amplitude of 3 $cm$ . Its maximum speed in $cm$/$s$ is

A particle executes linear simple harmonic motion with an amplitude of 2 cm. When the particle is at 1 cm from the mean position the magnitude of its velocity is equal to that of its acceleration. Then its time period in second is

The equation of SHM is given by $x = 3\sin 20\pi t + 4\cos 20\pi t$ Where x is in cm and t is I second. The amplitude is

$A1.00{\rm{ imes }}{10^{ - 20}}kg$ particle is vibrating with simple harmonic motion with a period of $1.00{\rm{ imes }}{10^{ - 5}}s$ and a maximum speed of $1.00 imes {10^3}m/s$ The maximum displacement of the particle is

The displacement of a particle varies according to the relation x=4(cos⁡πt+sin⁡πt). The amplitude of the particle is

The displacement of a particle executing SHM is given by $y = 0.25{\rm{ }}sin{\rm{ }}200t{\rm{ }}\,\,\,cm$ the maximum speed of the particle is

The acceleration of a particle in SHM is $0.8\,\,m{s^{ - 2}}$ when its displacement is 0.2 m. The frequency of oscillation is

If the displacement of a particle executing SHM is $y = 0.25\sin (200t + 0.8)m.$ its maximum velocity is