Maximum amplitude(in $cm$) of $SHM$ so block A will not slip on block $B , K =100 N / m$
AIIMS 2019, Diffcult
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The expression of force is given as,
$F = m \omega^{2} A =\mu mg$
So,
$A =\frac{\mu g }{\omega^{2}}$ And $\omega=\sqrt{\frac{ K }{ m }}$
So,
$\omega=\sqrt{\frac{100}{1.5}}$
Substitute the values.
$A =\frac{(0.4)(9.8)}{\left(\sqrt{\frac{100}{1.5}}\right)^{2}}$
$=6\, cm$
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