A body with speed '$v$' is moving along a straight line. At the same time it is at distance $x$ from a fixed point on the line, the speed is given by $v^2=144-9x^2$. Then:

Given:

$\begin{aligned} \quad \theta^{2} & =144-9 x^{2} \\ \therefore \quad \theta & =\sqrt{144-9 x^{2}} \\ & =\sqrt{9\left(16-x^{2}\right)} \quad \Rightarrow 3 \sqrt{16-x^{2}} \end{aligned}$

Comparing it with the velocity in Strm

$v=\omega \sqrt{A^{2}-x^{2}}$

Comparing the equation with (1)

$\begin{aligned} \therefore & =3 / \mathrm{s} \\ \text { and, } A^{2} & =16 \quad \therefore A=4 \mathrm{~m} \end{aligned}$

In SHM, displacement $<$ distance (as the particle passes mean position

- a correct making displacement $=0$

Now

$\begin{array}{l} a=-\omega^{2} x \quad(a \text {-accekration }) \\ w \text { - angular frequency } \\ x \text {-displacenent } \\ \therefore \text { at } x=3 \mathrm{~m}, \omega=3 \mathrm{~s}^{-1} \\ a=-(3)^{2}(3)=-27 \mathrm{~m} / \mathrm{s}^{2} \\ |a|=27 \mathrm{~m} / \mathrm{s}^{2} \text { (b correct) } \\ \end{array}$

Now $T=2 \pi / \omega \Rightarrow \frac{2 \bar{u}}{3}$ unit ( $c$ correct) The maximum displacement is une Amplitude $=4$ mit.

(Hence d correct)

Hence all ine above are correct.