A particle moves in a straight line with a velocity given by $v = \alpha t + \beta t^3$, where $\alpha$ and $\beta$ are constants. If its acceleration at $t=1$ second is twice its acceleration at $t=0$, what is the relation between $\alpha$ and $\beta$?

A particle is subjected simultaneously to two SHM’s one along the $x$-axis and the other along the $y$ -axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute

When the amplitude of a body executing SHM become twice what happens?

A particle executes linear SHM with an amplitude of $3$ cm. When the particle is at $2$ cm from the mean position, the magnitude of its velocity is equal to that of its acceleration. Then, its time period in seconds 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

A simple harmonic wave having an amplitude A and time period T is represented by the equation $y = 5{\rm{sin \pi }}\left( {t + 4} \right){\rm{m}}.$ Then the value of A in (metre) and T in (second) are

For a body in SHM the velocity is given by the relation $V = \sqrt {144 - 16{x^2}} \,\,m{s^{ - 1}}.$ The maximum acceleration is

A particle is subjected simultaneously to two SHM’s one along the $x$-axis and the other along the $y$ -axis. The two vibrations are in phase and have unequal amplitudes. The particle will execute

The displacement of a particle performing simple harmonic motion is given by, $x = 8\,\,\sin \,\,\omega t + 6\cos \omega t,$ where distance is in cm and time is in second. The amplitude of motion is

A body is vibrating in simple harmonic motion. If its acceleration is 12 ${\rm{cm}}{{\rm{s}}^{ - 2}}$ at a displacement 3 cm, then time period is

If the displacement of a particle executing SHM is given by $y = 0.30\,\,\sin \,\,\left( {220t + 0.64} \right)$〗 in metre, then the frequency and maximum velocity of the particle is