A block P of mass m is placed on a horizontal frictionless plane. A second block of same mass m is p

Q: Why do we consider the combined mass of both blocks to find the angular frequency?

Since the blocks oscillate together without slipping, they behave as a single rigid body of mass $2m$ attached to the spring.

Q: Why is the maximum acceleration used to find the maximum frictional force?

In SHM, acceleration varies continuously and reaches its maximum at the extreme positions. Since force is proportional to acceleration ($F = ma$), the maximum frictional force is required at these extreme points.

Q: Why does the spring force not act directly on block $P$?

The spring is physically connected only to block $Q$. Block $P$ is accelerated solely by the static friction force acting at the interface between $P$ and $Q$.

Q: What would happen if the amplitude $A$ is increased beyond the limit?

If the required frictional force exceeds the limiting static friction ($\mu_s mg$), block $Q$ will start slipping over block $P$, and they will no longer oscillate with the same frequency.

Q: Does the coefficient of static friction $\mu_s$ affect the maximum frictional force in this problem?

No, as long as there is no slipping, the actual static friction force is determined solely by the acceleration of the block ($f_s = ma$). The coefficient $\mu_s$ only determines the maximum limit of this force, not its instantaneous value.

Question

A block $P$ of mass $m$ is placed on a horizontal frictionless plane. A second block of same mass $m$ is placed on it and is connected to a spring of spring constant $k$, the two blocks are pulled by a distance $A$. Block $Q$ oscillates without slipping. What is the maximum value of frictional force between the two blocks?

Options:

(A) $\frac{kA}{2}$

(B) $kA$

(C) $\mu_s mg$

(D) Zero

Sigma Percentile · Accepted Answer

### The Magic of Co-oscillating Blocks

Imagine two blocks stacked on top of each other, sliding back and forth on a perfectly smooth, icy surface. The top block is connected to a spring, which pulls and pushes it relentlessly. Yet, the bottom block moves in perfect harmony with it, never slipping for a single moment. How does this happen? The answer lies in the invisible, silent partner of mechanics: **static friction**.

In this problem, we are challenged to find the maximum value of this frictional force during their simple harmonic journey. Let's dive deep into the physics of this stacked system and uncover the elegant mathematics that governs it.

### Analyzing the Stacked System

When the system is pulled by a distance $A$ and released, both blocks execute Simple Harmonic Motion (SHM) together. Because there is absolutely no slipping between block $Q$ and block $P$, they share the exact same acceleration at every single instant of time.

This crucial physical constraint allows us to treat the two blocks as a single, unified body of mass:

$$M = m + m = 2m$$

This combined mass is attached to a spring of force constant $k$. Therefore, the angular frequency $\omega$ of this co-oscillating system is given by the classic formula:

$$\omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{k}{2m}}$$

### The Peak of Acceleration

In any simple harmonic oscillator, the acceleration is not constant; it varies continuously as the system moves. The acceleration is zero at the mean position and reaches its maximum magnitude at the extreme positions, where the displacement equals the amplitude $A$:

$$a_{\max} = \omega^2 A$$

By substituting our expression for $\omega$ into this equation, we can find the maximum acceleration experienced by both blocks:

$$a_{\max} = \left(\sqrt{\frac{k}{2m}}\right)^2 A = \frac{kA}{2m}$$

This is the maximum acceleration that the system must sustain without slipping.

### Isolating the Lower Block

To find the frictional force between the blocks, we must perform a classic physics trick: **isolate one of the blocks** and draw its free-body diagram. Let's choose the lower block, block $P$.

Block $P$ rests on a frictionless horizontal plane, meaning the floor exerts no horizontal force on it. The only horizontal force acting on block $P$ is the static friction $f_s$ exerted by block $Q$ at their contact interface.

According to Newton's second law, this frictional force is solely responsible for accelerating block $P$:

$$f_s = m \cdot a$$

### Calculating the Maximum Friction

The maximum frictional force $f_{\max}$ is required at the exact instant when the acceleration of block $P$ is at its peak, $a_{\max}$:

$$f_{\max} = m \cdot a_{\max}$$

Substituting the value of $a_{\max}$ we calculated earlier:

$$f_{\max} = m \cdot \left(\frac{kA}{2m}\right)$$

Notice how beautifully the mass $m$ cancels out from the numerator and denominator! This leaves us with our final, elegant result:

$$f_{\max} = \frac{kA}{2}$$

This simple formula tells us that the maximum frictional force depends only on the spring constant $k$ and the amplitude $A$, and is completely independent of the mass of the blocks!

Select Answer:

Visualized Solution

Understanding the Two-Block System

The Combined System as a Single Oscillator

Determining Maximum Acceleration

Substituting $ω$ into $a_{m a x}$

Simplifying the Acceleration Expression

Isolating Block $P$ (Lower Block)

Substituting $a_{m a x}$ for Block $P$

Calculating the Maximum Frictional Force

Condition for No Slipping

The Sigma Insight: Simple Harmonic Motion (SHM)

The Magic of Co-oscillating Blocks

Analyzing the Stacked System

The Peak of Acceleration

Isolating the Lower Block

Calculating the Maximum Friction

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