Question

A particle is doing simple harmonic motion. Its maximum velocity is $3m/s$ and amplitude of motion is $0.2m.$ Calculate the period of motion.

Answer 1

Given $v_{max}=3m{s}^{-1};a=0.2nT=?$

$\because v_{max}={\omega }_{0}a\Rightarrow {\omega }_0=\frac{v_{max}}{a}$

$\therefore {\omega }_0=\frac{3}{0.2}=\frac{30}{2}=15$

or $\frac{2\pi }{T}=15$

$\therefore T=\frac{2\pi }{15}=\frac{6.28}{15}$

$=0.4186$

$T=0.42s$