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: \[A_{\max} = \frac{\mu (M+m) g}{k}\]

Key concepts

Static Friction

Static friction plays a crucial role when it comes to objects in contact that don't slip against each other, such as block B on block A in this setup. It is the force that keeps the two blocks from sliding over each other when the spring is in motion.
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- **Understanding Static Friction:** Static friction acts to resist the start of motion between two surfaces. Its magnitude depends on the coefficient of static friction (\(\mu\)) and the normal force, which in this case, is the gravitational force acting on block B (\(mg\)).

- **Formula:** The static friction force (\(f_s\)) can be written as \(f_s = \mu m g\). This represents the maximum force below which block B will not slide off block A. It is crucial because if the force applied exceeds this value, static friction can no longer hold the two blocks together.
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The role of static friction is fundamental in determining whether the block B stays on block A during the motion initiated by displacing the block-spring system. As the system moves, this concept is put to test at the maximum amplitude of oscillation, where static friction ensures that block B does not slide off.

Oscillation Amplitude

Oscillation amplitude refers to the maximum extent of displacement from the equilibrium position during the simple harmonic motion (S.H.M) of the mass-spring system. Here, it determines how far the system can move from the center before block B starts to slide off block A because the force would exceed the maximum static friction.
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- **Calculating Maximum Amplitude:** By analyzing the forces, we find that the spring force at maximum displacement equals the maximum static friction that can be developed. That's why the formula becomes essential: \[A_{\max} = \frac{\mu (M+m) g}{k}\]This formula gives the maximum permissible amplitude for the system without causing block B to slip.
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Understanding this concept is pivotal, as exceeding the amplitude means overcoming the static friction leading to sliding. Hence, correctly computing the maximum amplitude ensures that the SHM proceeds smoothly and the blocks remain intact. This emphasizes the critical balance between the oscillation's extent and the resistive static force present.

Mass-Spring System

The mass-spring system in this scenario describes a setup where two blocks are coupled with a spring that is allowed to oscillate along a horizontal frictionless surface. The system embodies the properties of simple harmonic motion (S.H.M), which is characterized by its sinusoidal properties in motion.
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- **System Dynamics:** The spring constant (\(k\)) measures the spring's resistance to deformation. When disturbed from an equilibrium position, the spring exerts an elastic restoring force proportional to the displacement (\(F = kx\)). This sets the system into oscillation around the equilibrium point.

- **Interplay of Forces:** In the context of the exercise, block A connects to the spring, and block B rests on block A. The forces include the gravitational pull, normal forces, and importantly, static friction between the blocks. As the spring elongates or compresses, it dictates the resulting motion characterized by the interplay of these forces.
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Ultimately, understanding the mass-spring system's functionality provides insight into the mechanics of S.H.M. By comprehending how it operates under varying conditions, such as different amplitudes and friction coefficients, one better appreciates how these parameters affect the stability and behavior of the system.