As shown in the figure, two blocks of masses $m_1$ and $\mathrm{m}_2$ are connected to a spring of force…

As shown in the figure, two blocks of masses $m_1$ and $\mathrm{m}_2$ are connected to a spring of force constant k . The blocks are slightly displaced in opposite directions to $\mathrm{x}_1$, $x_2$ distances and released. If the system executes simple harmonic motion, then the frequency of oscillation of the system ( $\omega$ ) is

1. $\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k^2$
2. $\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k^2}$
3. $\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right)}$
4. $\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k}$

Solution

$\mu=\frac{m_1 m_2}{m_1+m_2}$ $\therefore$ Frequency of oscillation is $\mathrm{W}=\sqrt{\frac{\mathrm{k}}{\mu}}=\sqrt{\frac{\mathrm{k}}{\left(\frac{\mathrm{~m}_1 \mathrm{~m}_2}{\mathrm{~m}_1+\mathrm{m}_2}\right)}}=\sqrt{\left(\frac{1}{\mathrm{~m}_1}+\frac{1}{\mathrm{~m}_2}\right) \mathrm{k}}$

Asked in: AP EAMCET 2024 (23 May Shift 1)

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