Two masses \[{m_1}\] and \[{m_2}\]are suspended by a massless spring of constant k. When the masses are in equilibrium \[{m_1}\] is removed without disturbing the system. Then the angular frequency of oscillation of \[{m_2}\] is
A. \[\sqrt {\dfrac{k}{{{m_1}}}} \]
B. \[\sqrt {\dfrac{k}{{{m_2}}}} \]
C. \[\sqrt {\dfrac{k}{{{m_1} + {m_2}}}} \]
D. \[\sqrt {\dfrac{k}{{{m_1}{m_2}}}} \]

Hint: Angular frequency of oscillation does not depend on the amplitude, it is directly proportional to the square root of force constant and inversely proportional to the square root of the mass.

Formula used:
\[\omega = \sqrt {\dfrac{k}{m}} \]

Complete step by step solution:
Two masses \[{m_1}\] and \[{m_2}\]are suspended by a spring of constant, when the masses are in equilibrium mass \[{m_1}\]is removed without disturbing the system we have to find the angular frequency of the system.

Image 1: Two blocks $m_1$ and $m_2$ suspended by spring.

Let the displacement be l when mass \[{m_1}\] and \[{m_2}\]and are suspended by the spring then for the system we have,
\[\left( {{m_1} + {m_2}} \right)g = kl\,.......(1)\]

Image 2: Force on the system when displacement is l.

Let the displacement be l’ when the mass is removed.
Similarly, we have,
\[{m_2}g = kl'\,.......(2)\]

As it is clear from equation (2) that only mass is responsible for the displacement when is removed without any disturbance in the system.

As the angular frequency is given by,
\[\omega = \sqrt {\dfrac{k}{m}} \,\]

Angular frequency when the mass is removed without any disturbance in the system will be,
\[\omega = \sqrt {\dfrac{k}{{{m_2}}}} \,\]

Hence, when the mass \[{m_1}\] will be removed angular frequency will only depend on the mass\[{m_2}\].

Therefore, option B is the correct option.

Note: As the spring is massless therefore all the mass-dependent quantities will only consist of masses of the blocks suspended with the spring.