Question

An object exceuting SHM has an acceleration of $0.2m/s^2$ at a distance of $0.05$m from the mean position calculate the period of SHM

• A. $2.1435$
• B. $3.1428$
• C. $1.432$
• D. $4.321$

Answer 1

The correct option is B $3.1428$

Given,

$a=0.2m/s^2$

$x=0.05m$

The time period of S.H.M is given by

$a={\omega }^{2}x$

$a=(\frac{2\pi }{T}{)}^{2}x$ ($\omega =\frac{2\pi }{T}$)

$T^2=4{\pi }^2\frac{x}{a}$

$T=2\pi \sqrt{\frac{x}{a}}$

$T=2\times 3.1428\times \sqrt{\frac{0.05}{0.2}}$

$T=3.1428sec$

The correct option is B.