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(D) For a two-body system connected by a spring,the equivalent mass (reduced mass) $\mu$ is given by:$\mu = \frac{m_1 m_2}{m_1 + m_2}$The angular freq

Vedclass · Accepted Answer

$\sqrt{\left(\frac{1}{m_1}+\frac{1}{m_2}\right) k}$

Vedclass · Answer

(D) For a two-body system connected by a spring,the equivalent mass (reduced mass) $\mu$ is given by:$\mu = \frac{m_1 m_2}{m_1 + m_2}$The angular frequency of oscillation $\omega$ is given by:$\omega = \sqrt{\frac{k}{\mu}}$Substituting the value of $\mu$:$\omega = \sqrt{\frac{k}{\left(\frac{m_1 m_2}{m_1 + m_2}\right)}}$$\omega = \sqrt{\frac{k(m_1 + m_2)}{m_1 m_2}}$$\omega = \sqrt{k \left(\frac{m_1}{m_1 m_2} + \frac{m_2}{m_1 m_2}\right)}$$\omega = \sqrt{k \left(\frac{1}{m_2} + \frac{1}{m_1}\right)}$Thus,the correct option is $D$.

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