The angular frequency is $\omega = 2\pi f$. For simple harmonic motion (SHM), velocity is $v = \pm \omega \sqrt{A^2 - x^2}$ and acceleration is $a = -\omega^2 x$.

[Q1] Maximum velocity occurs at $x=0$, and maximum acceleration occurs at $x=\pm A$.

$$v_{max} = A\omega = 2\pi f A$$ $$a_{max} = A\omega^2 = (2\pi f)^2 A$$

[Q2] At displacement $x$, the acceleration and velocity are:

$$a = -\omega^2 x = -(2\pi f)^2 x$$ $$v = \omega \sqrt{A^2 - x^2} = 2\pi f \sqrt{A^2 - x^2}$$

Substitute the given values: $f = 1000$ Hz, $A = 0.40 \times 10^{-3}$ m, and $x = 0.20 \times 10^{-3}$ m.

$v_{max} = 2\pi (1000)(0.40 \times 10^{-3}) = 0.8\pi$ m/s $a_{max} = (2\pi \times 1000)^2 (0.40 \times 10^{-3}) = 1600\pi^2$ m/s$^2$ $a = -(2\pi \times 1000)^2 (0.20 \times 10^{-3}) = -800\pi^2$ m/s$^2$ $v = 2\pi (1000) \sqrt{(0.40 \times 10^{-3})^2 - (0.20 \times 10^{-3})^2} = 2000\pi \times 10^{-3} \sqrt{0.40^2 - 0.20^2} = 0.4\pi\sqrt{3}$ m/s