Question

Values of the acceleration $A$ of a particle moving in simple harmonic motion as a function of its displacement $x$ are given in the table below.
The period of the motion is -

• A. $\frac{\pi }{2}s$
  
• B. $\frac{2}{\pi }s$
  
• C. $\pi s$
• D. $\frac{1}{\pi }s$
  

Answer 1

The correct option is C $\pi s$

As we know, for 𝑆𝐻𝑀 acceleration will be
$A=-{\omega }^{2}x$
Here, we took magnitude of a
$|A|={\omega }^{2}x$
$\frac{|A|}{x}={\omega }^2$
${\omega }^2=4$
$\omega =2$
As we know,
$\omega =\frac{2\pi }{T}$
$2=\frac{2\pi }{T}$
$T=\pi \sec$
Final answer: (d)