Q46 / 60 · JEE Advanced 2009 (Paper 1) · Physics
The $x-t$ graph of a particle undergoing simple harmonic motion is shown below. The acceleration of the particle at $t=\frac{4}{3} \mathrm{~s}$ is
- A$\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}$
- B$\frac{-\pi^2}{32} \mathrm{cms}^{-2}$
- C$\frac{\pi^2}{32} \mathrm{cms}^{-2}$
- Correct answerD$-\frac{\sqrt{3}}{32} \pi^2 \mathrm{cms}^{-2}$
Solution
$T=8 \mathrm{~s}, \omega=\frac{2 \pi}{T}=\left(\frac{\pi}{4}\right) \operatorname{rads}^{-1}$
$$
\begin{aligned}
x & =A \sin \omega t \\
\therefore \quad a=-\omega^2 x & =-\left(\frac{\pi^2}{16}\right) \sin \left(\frac{\pi}{4} t\right)
\end{aligned}
$$
Substituting $t=\frac{4}{3} \mathrm{~s}$, we get
$$
a=-\left(\frac{\sqrt{3}}{32} \pi^2\right) \mathrm{cms}^{-2}
$$
Hence, the correct answer is option: 'D'.