Question

The equation of motion of a particle executing simple harmonic motion is $a+16{\pi }^{2}x=0$ In this equation, $a$ is the linear acceleration in $m/s^2$ of the particle at a displacement $x$ in meter. The time period in simple harmonic motion is:

• A. $3\sec$
  
• B. $2\sec$
  
• C. $1\sec$
  
• D. $1/2\sec$

Answer 1

The correct option is D $1/2\sec$

Given,
𝑆𝐻𝑀 equation $=a+16{\pi }^{2}x=0$
As we know,
Acceleration of 𝑆𝐻𝑀 will be
$a=-{\omega }^{2}x$
Therefore,
$\omega =16{\pi }^2$
$\omega =4\pi$
As we know,
$T=2\pi /\omega$
Therefore,
$=\frac{2\pi }{4\pi }$
$=\frac{1}{2}\sec$
Final answer: (a)