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A block of mass $m$ , attached to a fixed position $O$ on a smooth inclined wedge of mass $M$ , oscillates with amplitude $A$ and linear frequency $f$ . The wedge is located on a rough horizontal surface. If the angle of the wedge is $60 °$ , then the force of friction acting on the wedge is given by (coefficient of static friction $= μ$ )

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$μ (M + m) ω^{2} A \sin ω t$

$\frac{1}{2} m ω^{2} A \sin ω t$

$μ (M + m) g$

$μ [(M + m) g + \frac{\sqrt{3}}{2} m ω^{2} A \sin ω t]$

answer is B.

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Detailed Solution

During the oscillation, the force exerted by the block along inclined plane

$F = m a = m (ω^{2} x) = m ω^{2} A \sin ω t$

Horizontal component of this will cause frictional force,so

$f = F \cos 60 ° = [m ω^{2} A \sin ω t] \cos 60 ° = \frac{1}{2} m ω^{2} A \sin ω t$

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