Question

Two equal masses are connected by a spring satisfying Hooke's law and are placed on a frictionless table. The spring is elongated a little and allowed to oscillate. Let the angular frequency of oscillations be $\omega$. Now one of the masses is stopped. The square of the new angular frequency is $\frac{{\omega }^2}{x}$, where $x$ is

Answer 1

Considering the frame of reference as centre of mass of two masses
$\sqrt{\frac{k}{\mu }}=\omega \mu =\frac{m_1{m}_2}{m_1+m_2}{m}_1=m_2=m\omega =\sqrt{\frac{2k}{m}}$
When one mass is stopped
${\omega }_2=\sqrt{\frac{k}{m}}{\omega }_2^2=\frac{k}{m}=\frac{{\omega }^2}{2}$