Question

Speed $v$ of a particle moving along a straight line, when it is at a distance $x$ from a fixed point on the line is given by
$v^2=108-9x^2$ (assuming mean position to have zero phase constant)
(all quantities in are in $cgs$ unit):

• A. The motion is uniformly accelerated along the straight line
• B. The magnitude of the acceleration at a distance $3cm$ from the fixed point is $27cm/s^2$
• C. The motion is simple harmonic about $x=\sqrt{12}m$.
• D. The maximum displacement from the fixed point is $4cm$.

Answer 1

Answer: (B)

The magnitude of the acceleration at a distance $3cm$ from the fixed point is $27cm/s^2$

$v^2=108-9x^2$ -----$\left(i\right)$

Differentiating $\left(i\right)$ w.r.t $x$

$2\times v\frac{dv}{dx}=-18x$ -----$\left(ii\right)$

We know that acceleration of particle is also given by,

$a$ = $v\frac{dv}{dx}$

From $\left(ii\right)$ we have ,

$2a=-18x$

hence ,

$a=-9x$

This acceleration is clearly not uniform.

At $x=3$

$a=-9\times 3$

$a=-2$

Hence the magnitude of acceleration at a distance of $3cm$ is $27cm/s^2.$