Q132 / 160 · AP EAMCET 2022 (06 Jul Shift 2) · Physics
An object is executing simple harmonic motion with an angular frequency $\omega$ If the maximum velocity is $v_{\max }$ then the maximum acceleration of the object is
- A$\omega^2 v_{\max }$
- Correct answerB$\omega v_{\max }$
- C$\omega \sqrt{v_{\max }}$
- D$3 \omega v_{\max }$
Solution
In SHM maximum velocity of object,
$$
\begin{aligned}
v_{\max } & =A \omega \\
\Rightarrow \quad A & =\frac{v_{\max }}{\omega}
...(i)\end{aligned}
$$
Where, $A$ is amplitude and $\omega$ is angular velocity. $\therefore$ Maximum acceleration of the object performing SHM,
$$
\begin{aligned}
& \alpha_{\max }=\omega^2 A \\
& =\omega^2 \frac{v_{\max }}{\omega} [From Eq. (i)]\\
& =\omega V_{\max }
\end{aligned}
$$
Hence, the correct answer is option: 'B'.