Question

An object of mans $m$ is undergoing linear $\text{SHM}$ according to $x=0.2\text{sin}\left(3.0t\right)\text{m}$. Find the velocity and acceleration of the mass when it is $+5\text{cm}$ from its mean position.

• A. $\pm 0.58\text{m/s},0.45{\text{m/s}}^2$
• B. $\pm 0.8\text{m/s},1.8{\text{m/s}}^2$
• C. $\pm 0.58\text{m/s},-0.45{\text{m/s}}^2$
• D. $\pm 0.8\text{m/s},0.45{\text{m/s}}^2$

Answer 1

The correct option is C $\pm 0.58\text{m/s},-0.45{\text{m/s}}^2$

Given,
$x=0.2\text{sin}\left(3.0t\right)\therefore A=0.2\text{m};\omega =3\text{rad/s}$
At $x=5\text{cm},$
$v=\pm \omega \sqrt{A^2-x^2}$
$=\pm 3\sqrt{(0.2{)}^2-(0.05{)}^2}$
$=\pm 0.58\text{m/s}$
$a=-{\omega }^{2}x=-\left(3{)}^2\right(0.05)$
$=-0.45{\text{m/s}}^2$