Question

A particle performing simple harmonic motion has maximum velocity $8cm/s$ and has acceleration $16cm/s^2$ at a distance of $4cm$ from the mean position. Calculate the time period and amplitude.

Answer 1

Given $v_{max}=8m{s}^{-1}$ and at $y=4cm$

$f=16cm{s}^{-2}$

$T=?;a=?$

$\because$ Maximum velocity $v_{max}={\omega }_{0}a$ or $8={\omega }_{0}a............\left(1\right)$

$\because$ Acceleration $f={\omega }_0^{2}y$

$\therefore 16={\omega }_0^2\times ..............\left(2\right)$

From eq (1) ${\omega }_0=\frac{8}{a}$

Substituting this value in eq (2)

$16=(\frac{8}{a}{)}^2\times 4=\frac{64}{a^2}\times 4$

$\therefore a^2=\frac{64\times 4}{16}=4\times 4$

$\therefore a=4cm$

$\because \omega )0=\frac{8}{a}=\frac{8}{4}=2$

or $\frac{2\pi }{T}=2\Rightarrow T=\pi =3.14s$