Two masses m_1 and m_2 are suspended together by a massless spring of constant K. When the masses are in equilibrium, m

Q: Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium, $m_1$ is removed without disturbing the system. The amplitude of oscillations is

a (a)With mass ${m_2}$ alone, the extension of the spring l is given as ${m_2}g = kl$ ...(i) With mass $({m_1} + {m_2})$, the extension $l'$ is given by $({m_1} + {m_2})g = k(l + \Delta l)$ ....(ii) The increase in extension is $\Delta l$ which is the amplitude of vibration. Subtracting (i) from (ii), we get ${m_1}g = k\Delta l$ or $\Delta l = \frac{{{m_1}g}}{k}$

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Two masses $m_1$ and $m_2$ are suspended together by a massless spring of constant $K$. When the masses are in equilibrium, $m_1$ is removed without disturbing the system. The amplitude of oscillations is

A$\frac{{{m_1}g}}{K}$
B$\frac{{{m_2}g}}{K}$
C$\frac{{({m_1} + {m_2})g}}{K}$
D$\frac{{({m_1} - {m_2})g}}{K}$

Medium

STD 11 Science
NEET
STD 11 - 13. oscillations
Physics

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