Answer
a) $a(t=0) = -271.4\mathrm{cm/s^2}$
b) $x(t) =(1.1\cos(15.7 t)-0.95\sin(15.7 t))\mathrm{cm}$
$v(t) =(-17.3 \sin(15.7 t)-15\cos(15.7 t))\mathrm{cm/s}$
$a(t) = (-271.4\cos(15.7 t)+235.6\sin(15.7 t))\mathrm{cm/s^2}$
Work Step by Step
Let the position of the needle be given by the general solution of the SHM
$x = A\cos(2\pi \nu t)+B\sin(2\pi \nu t)$
Here $\nu = 2.5\mathrm{Hz}$. By differentiating with respect to time, we get the velocity
$v = 2\pi\nu(-A\sin(2\pi \nu t)+B\cos(2\pi \nu t))$
At $t = 0$, $x = +1.1\mathrm{cm}\implies A = 1.1\mathrm{cm}$.
At $t = 0$, $v = -15\mathrm{cm/s} \implies 2\pi\nu B = -15\mathrm{cm/s}\implies B = -0.95\mathrm{cm}$
By differentiating $v$ by $t$, we get
$a = -(2\pi\nu)^2\times( A\cos(2\pi \nu t)+B\sin(2\pi \nu t)) = -\omega^2x$
where $\omega$ is the angular frequency, which is what we expect for a simple harmonic motion.
Plugging in the values for $A$ and $B$ and evaluating at $t=0$, we get
$a(t=0) = -(2\pi\nu)^2\times A = -271.4\mathrm{cm/s^2}$
The forms of $x$, $v$, and $a$ are simply obtained by putting in the numbers for the expressions above. We get
$x(t) =(1.1\cos(15.7 t)-0.95\sin(15.7 t))\mathrm{cm}$
$v(t) =(-17.3 \sin(15.7 t)-15\cos(15.7 t))\mathrm{cm/s}$
$a(t) = (-271.4\cos(15.7 t)+235.6\sin(15.7 t))\mathrm{cm/s^2}$
