Q.A particle performs linear S.H.M. starting from the mean position. Its amplitude is A and time period is T. At the instant when its speed is half the maximum speed, its displacement x is (A) $\frac{\sqrt3}{2}A$
(B) $\sqrt{\frac{2}{3}}A$
(C) $\frac{A}{\sqrt2}$
(D) $\frac{A}{2}$

For a particle in linear S.H.M., displacement x, velocity v and acceleration a are all determined at every instant by a single phase angle (omega t + phi): x = A sin(omega t + phi), v = Aomegacos(omega t + phi), a = -Aomega^2sin(omega t + phi) = -omega^2 x. Two starting conditions come up constantly: starting from the mean position (x = 0 at t = 0) gives phi = 0 and x = +/-A sin(omega t); starting from an extreme position (x = +/-A at t = 0) gives phi = pi/2 and x = +/-A cos(omega t).

The three quantities reach their extreme values at opposite points of the cycle: displacement and acceleration are BOTH zero at the mean position and BOTH maximum in magnitude (A and omega^2A respectively) at the extreme positions, while velocity is exactly out of step -- maximum (omegaA) at the mean position and zero at the extremes. Graphically (plotting x, v, a against phase or time), the three curves are all periodic, of the same period T; displacement and acceleration are always sine-type curves with a mutual phase difference of pi radian (exactly opposite in sign, consistent with a = -omega^2 x), while velocity is a cosine-type curve, a quarter-period (pi/2 radian) ahead of displacement and a further quarter-period ahead of acceleration.