Question

An object of mass 0.2 kg executes SHM along x-axis with a frequency of $\frac{25}{\pi }Hz$. At the position x=0.04 m the object has kinetic energy 0.5 J. The amplitude of oscillation will be:

• A. 0.06 m
• B. 0.04 m
• C. 0.05 m
• D. 0.25 m

Answer 1

The correct option is A 0.06 m

We know,

In SHM,

$v=\omega \sqrt{(A^2-x^2)}$

$KE=\frac{1}{2}m{v}^2=\frac{1}{2}m{\omega }^2(A^2-x^2)$

$0.5=0.5\times 0.2\times 4{\pi }^2\times \frac{{25}^2}{{\pi }^2}\times (A^2-{0.04}^2)$

$0.5=250(A^2-0.04)$

$A^2-0.0016=0.002$

$A^2=0.0036$

$A=0.06m$

Option $\text{A}$ is the correct answer