Q110 / 150 · MHT CET 2021 (20 Sep Shift 1) · Physics
A particle executes S.H.M. of period $\frac{2 \pi}{\sqrt{3}} \sec$ along a straight line $4 \mathrm{~cm}$ long. The displacement of the particle at which the velocity is numerically equal to the acceleration is
- A$2 \mathrm{~cm}$
- Correct answerB$1 \mathrm{~cm}$
- C$4 \mathrm{~cm}$
- D$3 \mathrm{~cm}$
Solution
Acceleration, $a=\omega^2 x$
Velocity, $v=\omega \sqrt{A^2-x^2}$
$$
\begin{aligned}
& \omega^2 \mathrm{x}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2} \\
& \omega^2 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \\
& 3 \mathrm{x}^2=\mathrm{A}^2-\mathrm{x}^2 \quad\left(\therefore \omega=\frac{2 \pi}{\mathrm{T}}=\sqrt{3}\right) \\
& 4 \mathrm{x}^2=\mathrm{A}^2 \\
& \mathrm{x}=\frac{\mathrm{A}}{2}=\frac{2}{2}=1 \mathrm{~cm}
\end{aligned}
$$
Hence, the correct answer is option: 'B'.