Question

The acceleration displacement graph of a particle executing simple harmonic motion is shown in fig. The time period of simple harmonic motion is

• A. $\frac{4 \pi}{\sqrt{3}} s$
• B. $\frac{2 \pi}{\sqrt{3}} s$
• C. The given graph doesn't represent simple harmonic motion.
• D. Information is insufficient.

Answer 1

Answer: (A)

$\frac{d a}{d x}=-\tan 37^{\circ}=-\frac{3}{4} \Rightarrow a=-\frac{3}{4} x$
so, $\omega^{2}=\frac{3}{2} \Rightarrow \frac{2 \pi}{n}=\frac{\sqrt{3}}{6} \Rightarrow T=\frac{4 \pi}{5} s$