Find the maximum amplitude (in cm) of the SHM | vedclass

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(C) For block $A$ not to slip on block $B$,the maximum pseudo force acting on $A$ must be less than or equal to the maximum static friction force.The

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$6$

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(C) For block $A$ not to slip on block $B$,the maximum pseudo force acting on $A$ must be less than or equal to the maximum static friction force.The total mass of the system is $M = m_A + m_B = 0.25 + 1.25 = 1.5 \ kg$.The angular frequency of the $SHM$ is $\omega = \sqrt{\frac{K}{M}} = \sqrt{\frac{100}{1.5}} = \sqrt{\frac{200}{3}} \ rad/s$.The maximum acceleration of the system is $a_{max} = \omega^2 A$.The condition for no slipping is $m_A a_{max} \leq \mu m_A g$,which simplifies to $a_{max} \leq \mu g$.Substituting $a_{max} = \omega^2 A$,we get $\omega^2 A \leq \mu g$.$A \leq \frac{\mu g}{\omega^2} = \frac{0.4 	imes 10}{100 / 1.5} = \frac{4}{100 / 1.5} = \frac{4 	imes 1.5}{100} = \frac{6}{100} \ m$.Converting to $cm$,$A = 6 \ cm$.

Find the maximum amplitude (in $cm$) of the $SHM$ such that block $A$ will not slip on block $B$. Given: spring constant $K = 100 \ N/m$,mass of block $A$ $m_A = 0.25 \ kg$,mass of block $B$ $m_B = 1.25 \ kg$,and coefficient of friction between $A$ and $B$ is $\mu = 0.4$. Take $g = 10 \ m/s^2$.

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