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(A) At the mean position, the velocity of mass $m_1$ is maximum, given by $v_{max} = \omega A = \sqrt{\frac{k}{m_1}} A$.When mass $m_2$ is placed on $

Vedclass · Accepted Answer

$[\frac{m_1 + m_2}{m_1}]^{1/2}$

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(A) At the mean position, the velocity of mass $m_1$ is maximum, given by $v_{max} = \omega A = \sqrt{\frac{k}{m_1}} A$.When mass $m_2$ is placed on $m_1$, the total mass becomes $(m_1 + m_2)$. Since the collision is perfectly inelastic and occurs at the mean position, the momentum is conserved.Initial momentum $P_i = m_1 v_{max}$.Final momentum $P_f = (m_1 + m_2) v'_{max}$, where $v'_{max}$ is the new maximum velocity.Since $P_i = P_f$, we have $m_1 v_{max} = (m_1 + m_2) v'_{max}$.Thus, $v'_{max} = \frac{m_1}{m_1 + m_2} v_{max}$.The new amplitude $A_1$ is given by $v'_{max} = \omega' A_1 = \sqrt{\frac{k}{m_1 + m_2}} A_1$.Equating the two expressions for $v'_{max}$:$\sqrt{\frac{k}{m_1 + m_2}} A_1 = \frac{m_1}{m_1 + m_2} \sqrt{\frac{k}{m_1}} A$.Simplifying, $A_1 = \sqrt{\frac{m_1 + m_2}{k}} \cdot \frac{m_1}{m_1 + m_2} \cdot \sqrt{\frac{k}{m_1}} A = \sqrt{\frac{m_1}{m_1 + m_2}} A$.Therefore, $A/A_1 = \sqrt{\frac{m_1 + m_2}{m_1}} = [\frac{m_1 + m_2}{m_1}]^{1/2}$.

$A$ mass $m_1$ performs $S.H.M.$ with amplitude $A$ while connected to a horizontal spring. While mass $m_1$ is passing through the mean position, another mass $m_2$ $(m_2 < m_1)$ is placed on it so that both masses move together with amplitude $A_1$. The ratio of $A/A_1$ is

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