Consider the following spring-blocks system. The coefficient of friction between blocks of masses m and 2m is $μ = 2 \tan θ$ . There is no friction between block of mass 2m and inclined plane. Maximum amplitude of the system along the incline about the equilibrium position for which there is no relative motion between the blocks is:

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$\frac{m g \sin θ}{k}$

$g \sin θ \sqrt{\frac{k}{m}}$

$\frac{3 m g \sin θ}{k}$

$g \cos θ \sqrt{\frac{k}{m}}$

answer is C.

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Detailed Solution

Angular frequency of SHM, $ω = \sqrt{\frac{k}{3 m}}$

$The maximum static frictional force is f_{\max} = μmgcosθ = 2 tanθmgcosθ = 2 mgsinθ$

$Applying newton's second law on the upper block at lower extreme position,$

$f - mgsinθ = {mω}^{2} A$

$\Rightarrow f = {mω}^{2} A + mgsinθ$

$As f \leq f_{\max},$

$\Rightarrow ω^{2} A \leq gsinθ$

$\Rightarrow A \leq \frac{3 mgsinθ}{k}$

Thus, $A_{\max} = \frac{3 mgsinθ}{k}$

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