Two blocks of mass 10 kg and 2 kg are connected by an ideal spring of spring constant K = 800 N/m and the system is placed on a horizontal surface as shown.
The coefficient of friction between 10 kg block and surface is 0.5 but friction is absent between 2 kg and the surface. Initially blocks are at rest and spring is relaxed. The 2 kg block is displaced to elongate the spring by 1 cm and is then released.
Then

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10 kg block executes SHM with time period $\frac{π}{5} s$

10 kg block will not move

Time period of oscillation of the system is $\frac{π}{10} s$

$\frac{π}{5} s$

answer is B, C.

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Detailed Solution

(a) The maximum friction force that can act on 10 kg block is $μ$ mg = 0.5 × 10 × 10 = 50 N
Maximum spring force = kx = 800 × 0.01 = 8N
The friction acting on 10 kg block is large enough to prevent its slipping. So it will not move.
(b) The 2 kg block will perform SHM. Equation of it motion will be – x = A cos ( $ω$ t)
Where $ω = \sqrt{\frac{K}{m}} and A = 0 .01 m$
So T =2 $π \sqrt{\frac{2}{800}} s = \frac{π}{10} s$