Question

The 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 $v^2=144–9x^2.$ Select the correct alternative(s) :

• A. The motion of the particle is SHM with time period T= $\frac{2\pi }{3}$ unit
• B. The maximum displacement of the particle from the mean position is 4 unit
• C. The magnitude of acceleration at a distance 3 units from the mean position is 27 unit
• D. The motion of the particle is periodic but not simple harmonic

Answer 1

Answer: (A), (B), (C)

The magnitude of acceleration at a distance 3 units from the mean position is 27 unit

Given
$v^2=144-9x^2$
$2v\frac{dv}{dx}=-18x$
$\therefore v\frac{dv}{dx}=-9x\Rightarrow \omega =3$
$\Rightarrow T=\frac{2\pi }{3}$ units
Also, $v^2=144-9x^2$
When $v=0$
$x^2=\frac{144}{9}\Rightarrow x=\frac{12}{3}$
$x=4$ units
Now, $\left|\begin{array}{c}a\end{array}\right|={\omega }^{2}x=9\times 3=27$ units