Question

An object of mass $0.2$kg executes simple harmonic oscillations along the x-axis with a frequency of $\left(25/\pi \right)$Hz. At the position $x=0.04$m, the object the kinetic energy of $0.5$ J and potential energy of $0.4$J. The amplitude of oscillations is _________m.

• A. 0.06 m
• B. 6 m
• C. 600 m
• D. 60 m

Answer 1

The correct option is A 0.06 m

Total Energy $=$ Kinetic energy $+$ Potential Energy

$T.E=K.E+P.E$

$⟹T.E=0.5$ $J+0.4$ $J$

$⟹T.E=0.9$ $J$

$T.E=\frac{1}{2}m{\omega }^2{a}^2[\omega =2\pi f(f\to frequencyanda\to amplitude\left)\right]$

$⟹0.9=\frac{1}{2}\times 0.2\times {(2\pi \times \frac{25}{\pi })}^2\times a^2$

$a=\sqrt{\frac{0.9}{[\frac{1}{2}\times 0.2\times (2\pi \times {\frac{25}{\pi }}^2)]}}$

$⟹a=0.06$ $m$