Question

A body of mass $0.1kg$ is executing simple harmonic motion according to the equation
$x=0.5\cos (100t+\frac{3\pi }{4})$ metre. Find: (i) the frequency of oscillation, (ii) initial phase, (iii) maximum velocity, (iv) maximum acceleration, (v) total energy.

Answer 1

The equation of motion of S.H.M $x=0.5\cos (100+\frac{3\pi }{4})$. Comparing it with general equation motion of S.H.M, described as follows, we get, $x=A\cos (\omega t+\theta )$

$\left(1\right)$ Frequency$=\frac{\omega }{2\pi }$, Where $\omega =100$

$=\frac{100}{2\pi }=\left(50\pi \right)Hz$

$\left(2\right)$Initial phase$\varphi =\frac{3\pi }{4}$

$\left(3\right)$Amplitude $A=0.5cm$

Maximum velocity$=\omega A=100\times 0.5=500m/s$

$\left(4\right)$Maximum acceleration$={\omega }^{2}A={100}^2\times 0.5=50m/s^2$

$\left(5\right)$Total energy$=\frac{1}{2}m{\omega }^2{A}^2$(Mass of the particle$=0.1kg$)

$=\frac{1}{2}\times 0.1\times (100{)}^2\times (0.5{)}^2$

$=125J$