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For the section b

In the given problem, I can understand that after placing the two blocks in equilibrium it oscillates with an amplitude,

$$A=\frac{2}{k}(m_1+m_2)g\sin\phi$$

The answer for (b) is given as

$$\frac{1}{k}(m_1+m_2)g\sin\phi$$

To my knowledge, $m_2$ will separate from $m_1$ when the acceleration is greater than $g\sin\phi$ and so they should be separating only at max displacement on the right side as the $m_1$ retraces its path after imparting the greatest acceleration to $m_2$. So shouldn't the answer be $\frac{2}{k}(m_1+m_2)g\sin\phi$ on the right? Where am I going wrong?

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  • $\begingroup$ It asks for the speed $\endgroup$
    – bemjanim
    Commented Dec 13, 2019 at 7:55
  • $\begingroup$ No, (b)- it asks the position where the m2 is released $\endgroup$
    – Vivek karunakaran
    Commented Dec 13, 2019 at 8:02
  • $\begingroup$ oh yeah decent, your answer looks right $\endgroup$
    – bemjanim
    Commented Dec 13, 2019 at 8:20
  • $\begingroup$ I believe the position is measured from the initial decompressed position (without masses) of the spring, not from the maximum compressed one. $\endgroup$
    – user65081
    Commented Dec 13, 2019 at 18:07
  • $\begingroup$ Try to calculate the constraint force (acceleration free) between m1 and m2 , put it equal zero and calculate the position $\endgroup$
    – Eli
    Commented Dec 13, 2019 at 20:39

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