Question

For a particle of mass $m$ executing SHM with angular frequency $\omega $ , the kinetic energy $k$ is given by $k=k_{0}cos^{2} \omega t .$ The equation of its displacement can be

• A. $\left(\frac{k_0}{m \omega^2}\right)^{1 / 2} \sin \omega t$
• B. $\left(\frac{2 k_0}{m \omega^2}\right)^{1 / 2} \sin \omega t$
• C. $\left(\frac{2 \omega^2}{m k_0}\right)^{1 / 2} \sin \omega t$
• D. $\left(\frac{2 k_0}{m \omega}\right)^{1 / 2} \sin \omega t$

Answer 1

Answer: (B)

If $m$ is the mass, $r$ is the amplitude of oscillation, then maximum kinetic energy,
$k_0=\frac{1}{2} m \omega^2 r^2 \quad$ or $\quad r=\left(\frac{2 k_0}{m \omega^2}\right)^{\frac{1}{2}}$
The displacement equation can be $y=r \sin \omega t=\left(\frac{2 k_0}{m \omega^2}\right)^{\frac{1}{2}} \sin \omega t$