A highly rigid cubical block A of small mass M and side L is fixed rigidly on to another cubical blo

Q: Why do we only consider the mass of block $A$ and ignore the mass of block $B$ in the time period calculation?

The problem states that block $A$ has a small mass $M$, but implicitly assumes block $B$ is either massless or its mass is negligible compared to $M$ for the oscillation dynamics. If block $B$'s mass were significant, we would have to account for its kinetic energy, which would require a more complex continuous-mass analysis.

Q: How did we get the shear strain as $\theta = x/L$?

Shear strain is defined as the tangent of the angle of deformation, $\tan\theta$. For a block of height $L$ shifted horizontally by $x$, $\tan\theta = x/L$. Since the displacement $x$ is very small, we can use the small-angle approximation $\theta \approx \tan\theta = x/L$.

Q: Why is the area of the sheared surface $L^2$ instead of $L^3$?

Area is a two-dimensional quantity. The shear stress acts parallel to the upper face of block $B$, which is a square of side $L$. Therefore, the area of this face is $A = L \times L = L^2$.

Q: What is the physical meaning of modulus of rigidity $\eta$?

The modulus of rigidity (or shear modulus) measures a material's resistance to shear deformation. A high $\eta$ means the material is very stiff and hard to slide sideways (like steel), while a low $\eta$ means it is easily sheared (like rubber or jelly).

Q: Can we use the spring mass equivalence $T = 2\pi \sqrt{M/k}$ directly here?

Yes! Once you show that the restoring force is of the form $F = -k_{\text{eff}} x$, where $k_{\text{eff}} = \eta L$, you can directly substitute this effective spring constant into the standard time period formula $T = 2\pi \sqrt{M/k_{\text{eff}}}$ to get the same result.

Question

A highly rigid cubical block $A$ of small mass $M$ and side $L$ is fixed rigidly on to another cubical block $B$ of the same dimensions and of low modulus of rigidity $\eta$ such that the lower face of $A$ completely covers the upper face of $B$. The lower face of $B$ is rigidly held on a horizontal surface. A small force $F$ is applied perpendicular to one of the side faces of $A$. After the force is withdrawn, block $A$ executes small oscillations, the time period of which is given by

Options:

(A) $2\pi \sqrt{M\eta L}$

(B) $2\pi \sqrt{\frac{M\eta}{L}}$

(C) $2\pi \sqrt{\frac{ML}{\eta}}$

(D) $2\pi \sqrt{\frac{M}{\eta L}}$

Sigma Percentile · Accepted Answer

### The Dance of Rigidity and Elasticity

Imagine a massive, highly rigid block sitting on top of a softer, jelly-like block. When you push the top block, it doesn't bend or warp; it simply slides as a single unit. But the block underneath—the elastic one—shears, stretching like a deck of cards being slid sideways. This beautiful interplay between rigid mass and elastic deformation is the heart of our problem.

In this classic JEE problem from 1992, we are asked to find the time period of small oscillations of a rigid block $A$ resting on a deformable block $B$. This is not just a math exercise; it is a fundamental exploration of how **elastic restoring forces** drive **simple harmonic motion (SHM)**.

Let's dive deep into the physics of shear deformation and discover how the geometry of the blocks dictates the rhythm of their oscillation.

### Analyzing the Setup

Let's look at the two blocks. Block $A$ has a mass $M$ and side $L$. It is highly rigid, which means we can treat it as a point mass concentrated at its center of gravity when it comes to translation. It doesn't deform; it only moves.

Block $B$, however, is the elastic engine of this system. It has the same dimensions (side $L$) but a low modulus of rigidity $\eta$. Its bottom face is glued to a horizontal table, while its top face is glued to the bottom of block $A$.

When we apply a horizontal force $F$ to block $A$, it shifts by a small distance $x$. Because block $A$ is glued to block $B$, the top face of block $B$ also shifts by $x$, while its bottom face remains fixed. This creates a **shear strain** in block $B$.

### The Physics of Shear: Modulus of Rigidity

To understand how block $B$ fights back against this displacement, we must look at the **modulus of rigidity** $\eta$. By definition, the modulus of rigidity is the ratio of shear stress to shear strain:

$$\eta = \frac{	ext{Shear Stress}}{	ext{Shear Strain}}$$

Shear stress is the restoring force $F$ per unit area $A$ of the face parallel to the force:

$$	ext{Shear Stress} = \frac{F}{A}$$

Since block $B$ is a cube of side $L$, the area of its top face is:

$$A = L^2$$

Shear strain is the angle of deformation $	heta$. For a small horizontal displacement $x$ and height $L$, we can use the small-angle approximation:

$$	heta \approx 	an	heta = \frac{x}{L}$$

### Finding the Restoring Force

Now, let's substitute these geometric relations back into our definition of $\eta$:

$$\eta = \frac{F / L^2}{x / L}$$

Simplifying this fraction, we get:

$$\eta = \frac{F}{L x}$$

Solving for the magnitude of the restoring force $F$, we find a remarkably simple relation:

$$F = \eta L x$$

Notice how this force is directly proportional to the displacement $x$! This is exactly like Hooke's Law for a spring, where the effective spring constant is:

$$k_{	ext{eff}} = \eta L$$

This is a beautiful realization: a sheared elastic block behaves exactly like a horizontal spring with a stiffness proportional to its modulus of rigidity and its linear dimension!

### The Equation of Motion and Time Period

Since the restoring force always acts in the direction opposite to the displacement, we can write the equation of motion using Newton's second law:

$$F_{	ext{restoring}} = -\eta L x$$

$$M a = -\eta L x$$

Dividing by the mass $M$, we get the acceleration $a$ of block $A$:

$$a = -\left(\frac{\eta L}{M}ight) x$$

This is the standard differential equation of simple harmonic motion, which has the form:

$$a = -\omega^2 x$$

By comparing the two equations, we can immediately identify the angular frequency $\omega$ of the oscillation:

$$\omega^2 = \frac{\eta L}{M} \implies \omega = \sqrt{\frac{\eta L}{M}}$$

### The Final Rhythm

The time period $T$ of the oscillation is the time taken for one complete cycle, given by:

$$T = \frac{2\pi}{\omega}$$

Substituting our value of $\omega$:

$$T = 2\pi \sqrt{\frac{M}{\eta L}}$$

This is our final elegant result, which corresponds perfectly to **Option (d)**.

### A Deeper Intuition

Let's look at the formula we derived: $T = 2\pi \sqrt{\frac{M}{\eta L}}$. Does it make physical sense?

First, if the mass $M$ of block $A$ increases, the time period $T$ increases. This is highly intuitive: a heavier block has more inertia, making it harder to accelerate, so it oscillates more slowly.

Second, if the modulus of rigidity $\eta$ increases, the material becomes stiffer, the restoring force becomes stronger, and the time period $T$ decreases. The block oscillates faster!

Third, if the side $L$ increases, the time period $T$ decreases. Why? Because a larger contact area $A = L^2$ increases the restoring force faster than the increased height $L$ reduces the shear strain. Thus, a larger block is stiffer in shear!

This problem is a masterclass in connecting the microscopic elastic properties of materials to macroscopic oscillatory motion. By mastering these connections, you unlock the true power of physics!

Select Answer:

Visualized Solution

Understanding the Physical Setup

Defining Modulus of Rigidity $η$

Expressing Area $A$ and Shear Angle $θ$

Formulating the Restoring Force

Finding the Acceleration of Block $A$

Comparing with the Standard SHM Equation

Calculating the Time Period $T$

Exploring Further: What if Block $A$ is also Deformable?

The Sigma Insight: Simple Harmonic Motion (SHM)

The Dance of Rigidity and Elasticity

Analyzing the Setup

The Physics of Shear

Finding the Restoring Force

The Equation of Motion and Time Period

The Final Rhythm

A Deeper Intuition

Similar Questions

Two identical springs of spring constant $2 k$ are attached to a block of mass $m$ and to fixed support (see figure). When the mass is displaced from equilibrium position on either side, it executes simple harmonic motion. The time period of oscillations of this system is

A wooden cube (density of wood $d$ ) of side $l$ floats in a liquid of density $ρ$ with its upper and lower surfaces horizontal. If the cube is pushed slightly down and released, it performs simple harmonic motion of period, $T$ . Then, $T$ is equal to

In the given figure, a body of mass $M$ is held between two massless springs, on a smooth inclined plane. The free ends of the springs are attached to firm supports. If each spring has spring constant $k$ , then the frequency of oscillation of given body is