Two masses m₁ and m₂ are supended together by a massless spring of constant k . When masses are

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13.Oscillations

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Two masses $m_1$ and $m_2$ are supended 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 vibration is

A

$m_1g / k$

B

$m_2g / k$

C

$\frac{{\left( {{m_1} + {m_2}} \right)\,g}}{k}$

D

$\frac{{\left( {{m_2} - {m_1}} \right)\,g}}{k}$

Solution

In equillibrium $F=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{g}$

By removing mass $m_1$

$F^{\prime}=\mathrm{m}_{2} \mathrm{g}$

$F_{\mathrm{r}}=F^{\prime}-F=\mathrm{m}_{2} \mathrm{g}-\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right)$

$\mathrm{kx}=\mathrm{m}_{1} \mathrm{g} \Rightarrow \mathrm{x}=\frac{\mathrm{m}_{1} \mathrm{g}}{\mathrm{k}}$

Standard 11

Physics

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