Question

As shown in figure, a block A having mass \(M\) is attached to one end of a massless spring. The block is on a frictionless horizontal surface and the free end of the spring is attached to a wall. Another block B having mass ' \(\mathrm{m}\) ' is placed on top of block A. Now on displacing this system horizontally and released, it executes S.H.M. What should be the maximum amplitude of oscillation so that B does not slide off A? Coefficient of static friction between the surfaces of the block's is \(\mu\). (A) \(A_{\max }=\\{(\mu \mathrm{mg}) / \mathrm{k}\\}\) (B) \(A_{\max }=[\\{\mu(m+M) g\\} / k]\) (C) \(A_{\max }=[\\{\mu(M-\mathrm{m}) g\\} / \mathrm{k}]\) (D) \(A_{\max }=[\\{2 \mu(M+m)\\} / k]\)

Short answer

The maximum amplitude of oscillation so that block B does not slide off block A is given by: \[A_{\max} = \frac{\mu (M+m) g}{k}\]

Step-by-step solution

Analyze the forces in system

To find the maximum amplitude, we must first analyze the forces acting on the blocks during the S.H.M, specially when the blocks reverses its direction. At maximum displacement, the force experienced by both blocks is equal to \(mg\), the gravitational force. Consider the forces acting on block B: gravitational force \(mg\) and static friction force \(f_s\). Since the maximum acceleration of block B is given by the static friction force, we write: \[f_s = ma \Rightarrow \mu m g = m a\] Now, let's analyze the forces acting on block A: for this block, the total force is given by the elastic force of the spring (\(kA\)) minus the static friction force on A: \[kA - f_s = MA \Rightarrow kA - \mu m g = M a\]

Solve for acceleration

To solve for the acceleration of the system, we can substitute the acceleration of block A from the equation we got in Step 1: \[kA - \mu m g = M (\frac{\mu m g}{m}) \Rightarrow kA = \mu (M+m) g\]

Solve for maximum amplitude

Now that we have the equation for the elastic force, we can easily solve for the maximum amplitude by dividing both sides of the expression by \(k\): \[A_{\max} = \frac{\mu (M+m) g}{k}\] Comparing this expression to the given options, it matches with option (B). Therefore, the maximum amplitude of oscillation so that block B does not slide off block A is given by: \[A_{\max} = \frac{\mu (M+m) g}{k}\]