Question

An object of mass 0.2 kg executes simple harmonic motion along $x$-axis with frequency of 25/$\pi$ Hz. At the position $x$ = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy of 0.4 J. The amplitude of oscillation is equal to

• A. 0.05 m
• B. 0.06 m
• C. 0.01 m
• D. 0.02 m

Answer 1

Answer: (B)

Total energy, $E = \frac{1}{2} m \omega^2 A^2$
$E = \frac{1}{2} m ( 2\pi \upsilon)^2 A^2 \:\:\:\: (\because \, \omega = 2 \pi \upsilon)$
$ \therefore \, A = \frac{1}{2 \pi \upsilon} \sqrt{\frac{2E}{m}}$
Putting $E = K + U ,$ we get
$A = \frac{1}{2\pi\left(\frac{25}{\pi}\right)}\sqrt{\frac{2\times\left(0.5+0.4\right)}{0.2}}= $ 0.06 m