Two masses m_1 and m_2 are suspended together | vedclass

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Question

(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 giv

Vedclass · Accepted Answer

$\frac{{{m_1}g}}{K}$

Vedclass · Answer

(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 $\Delta l = \frac{{{m_1}g}}{k}$

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

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