Question 37: The x – t graph of a particle performing simple harmonic motion is shown in the figure. The acceleration of the particle at t=2 \mathrm{~s} is :
(1) \frac{\pi^{2}}{16} \mathrm{~m} \mathrm{~s}^{-2}
(2) -\frac{\pi^{2}}{16} \mathrm{~m} \mathrm{~s}^{-2}
(3) \frac{\pi^{2}}{8} \mathrm{~m} \mathrm{~s}^{-2}
(4) -\frac{\pi^{2}}{8} \mathrm{~m} \mathrm{~s}^{-2}
Answer: Option (2)
Explanation:
From the given x–t graph, the maximum displacement of the particle is 1 \mathrm{~m}, which is the amplitude.
The particle reaches maximum displacement at t=2 \mathrm{~s}.
In simple harmonic motion, maximum displacement occurs at one-fourth of the time period.
Hence, the time period is T=4 \times 2 = 8 \mathrm{~s}.
The angular frequency is given by \omega=\frac{2\pi}{T}.
Substituting T=8, we get \omega=\frac{2\pi}{8}=\frac{\pi}{4} \mathrm{~rad\,s^{-1}}.
In simple harmonic motion, acceleration is given by a=-\omega^{2} x.
At t=2 \mathrm{~s}, the displacement is x=+1 \mathrm{~m}.
Therefore, a=-\left(\frac{\pi}{4}\right)^{2} \times 1.
So, a=-\frac{\pi^{2}}{16} \mathrm{~m} \mathrm{~s}^{-2}.
Hence, the correct answer is Option (2).