A mass attached to a spring performs S.H.M. whose displacement is $\mathrm{x}=3 \times 10^{-3} \cos 2 \pi \mathrm{t}$ metre. The time taken to obtain maximum speed for the first time is

The displacement of a mass in simple harmonic motion is given by $x = 3 \times 10^{-3} \cos(2\pi t)$ meters.

This matches the standard form $x = A \cos(\omega t)$, where $A = 3 \times 10^{-3}$ m is the amplitude and $\omega = 2\pi$ rad/s is the angular frequency.

Velocity is found by differentiating displacement: $v = \frac{dx}{dt} = -6\pi \times 10^{-3} \sin(2\pi t)$ m/s. Speed reaches its maximum when $|\sin(2\pi t)| = 1$, which occurs as the mass crosses the equilibrium position for the first time after starting at maximum positive displacement.

The time of maximum speed corresponds to $\cos(2\pi t) = 0$. The smallest positive solution is $2\pi t = \frac{\pi}{2}$, giving $t = \frac{1}{4}$ s.

Final answer: $t = \frac{1}{4}$ s.