Question

If displacement $x$ of a moving particle in a plane at time $t$ is given by $x=2\cos \left(t\sqrt{3}\right)+3\sin \left(t\sqrt{3}\right)$, then its acceleration is proportional to

• A. $\frac{1}{x}$
• B. $x$
• C. $x^2$
• D. $\frac{1}{x^2}$

Answer 1

The correct option is B

$x$

Step 1: Given data

Given, $x=2\cos \left(t\sqrt{3}\right)+3\sin \left(t\sqrt{3}\right)$

Where $x$ is displacement and $t$ is time.

Step 2: Calculating velocity

$x=2\cos \left(t\sqrt{3}\right)+3\sin \left(t\sqrt{3}\right)$

Velocity,

$$\begin{gathered} v=\frac{dx}{dt} \\ v=\frac{d\left(2\cos \left(t\sqrt{3}\right)+3\sin \left(t\sqrt{3}\right)\right)}{dt} \\ v=-2\sqrt{3}\sin \left(t\sqrt{3}\right)+3\sqrt{3}\cos \left(t\sqrt{3}\right) \end{gathered}$$

Step 3: Calculating acceleration

Acceleration,

$$\begin{gathered} a=\frac{dv}{dt} \\ a=\frac{d\left(-2\sqrt{3}\sin \left(t\sqrt{3}\right)+3\sqrt{3}\cos \left(t\sqrt{3}\right)\right)}{dt} \\ a=-6\cos \left(t\sqrt{3}\right)-9\sin \left(t\sqrt{3}\right) \\ a=-3\left(2\cos \left(t\sqrt{3}\right)+3\sin \left(t\sqrt{3}\right)\right) \\ a=-3x \end{gathered}$$

$\therefore a\propto x$

Acceleration is proportional to displacement.

Hence, option B is correct