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 given by $v^2=144-9x^2$.
Choose the wrong option.

• A. $\text{Displacement of the particle}\le \text{distance moved by it}$
• B. $\text{The magnitude of acceleration at a distance 3 units from the fixed point is 27 units}$
• C. $\text{The motion is simple harmonic with}T=\frac{\pi }{3}\text{units}$
• D. $\text{The maximum displacement from the fixed point is 4 units}$

Answer 1

The correct option is C $\text{The motion is simple harmonic with}T=\frac{\pi }{3}\text{units}$

Given,
$v^2=144-9x^2$
Comparing this with
$v=\omega \sqrt{A^2-x^2}$
$v^2={\omega }^2{A}^2-{\omega }^2{x}^2$
we get,
${\omega }^2=9\Rightarrow \omega =\pm 3$
and
${\omega }^2{A}^2=144\Rightarrow A=4\text{units}$
So, option (d) is true.
We know that, $|a|={\omega }^{2}x\Rightarrow |a|=\left(9\right)\left(3\right)=27\text{units}$
$\because \omega =\frac{2\pi }{T}$ we get , $T=\frac{2\pi }{3}\text{units}$
So, option (b) is true and option (c) is wrong.
Since, statement (a) is always correct.
we can say that, option (c) is wrong answer.