Question

A particle executes simple harmonic motion with an amplitude of 5 cm. when the particle is at 4 cm from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. then, its periodic time in second is:

• A. $\frac{7}{3}\pi$
• B. $\frac{3}{8}\pi$
• C. $\frac{4\pi }{3}$
• D. $\frac{8\pi }{3}$

Answer 1

The correct option is D $\frac{8\pi }{3}$

We know that velocity in S. H. M. is given by

$v=\omega \sqrt{a^2-x^2}$

$v=\omega \sqrt{5^2-4^2}=3\omega$

$a={\omega }^2\times 4$

Given |A| = |v|

$4{\omega }^2=3\omega$

$\omega =\frac{3}{4}=\frac{2H}{T}$

$T=\frac{8H}{3}$ sec.