Question

A $2.0kg$ particle undergoes SHM according to $x=1.5\sin (\frac{\pi t}{4}+\frac{\pi }{6})$
(in SI units). What is the total mechanical energy of the particle?

Answer 1

$E=\frac{1}{2}m{v}_{max}^2=\frac{1}{2}\left(m\right){\omega }^2{A}^2$

$\because v=\omega A$

Also, $\omega =\frac{\pi }{4}$ and $A=1.5$

$E=\frac{1}{2}\times 2.0\times {\left(\frac{\pi }{4}\right)}^2\times {\left(1.5\right)}^2$

$=1.39J$