One end of a light spring of force constant k is fixed to a block A of mass M placed on a horizontal frictionless table; the other end of the spring is fixed to a wall (Fig). A smaller block B of mass m is placed on block A. The system is displaced by a small amount and released. What is the maximum amplitude of the resulting simple harmonic motion of the system so that the upper block does not slip over the lower block? The coefficient of static friction between the two blocks is $μ$ .

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$A_{\max} = \frac{μ (M + m) g}{k}$

$A_{\max} = \frac{μ m g}{k}$

$A_{\max} = \frac{μ (M - m) g}{k}$

$A_{\max} = \frac{μ M g}{k}$

answer is C.

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Detailed Solution

The angular frequency of the system is
$ω = {[\frac{k}{(M + m)}]}^{1 / 2}$ …………(1)
The upper block of mass m will not slip over the lower block of mass M if the maximum force on the upper block $f_{\max}$ does not exceed the frictional force $μ m g$ between the two blocks. Now
$f_{\max} = m a_{\max} = m ω^{2} A_{\max}$ ………..(2)
where $a_{\max}$ is the maximum acceleration and $A_{\max}$ -is the maximum amplitude. Using (1) in (2), we get
$f_{\max} = \frac{m k A_{\max}}{(M + m)}$
For no slipping, $f_{\max} = μ m g$
or $\frac{m k A_{\max}}{(M + m)} = μ m g o r A_{\max} = \frac{μ (M + m) g}{k}$ .