A massless rod BD is suspended by two identical massless strings AB and CD of equal lengths. A block

Q: Why does the fundamental frequency depend on tension?

Tension determines the speed of transverse waves traveling along the string ($v = \sqrt{T/\mu}$). Since the wavelength of the fundamental mode is fixed by the string's length ($\lambda = 2L$), a higher wave speed directly results in a higher fundamental frequency ($f = v/\lambda$).

Q: Why did we choose the pivot at point P instead of B or D?

Choosing the pivot at point P is a strategic choice because the downward force of the hanging mass ($mg$) acts directly at P. This makes its torque zero, allowing us to write a torque equation containing only the tensions $T_{AB}$ and $T_{CD}$, completely eliminating the unknown mass $m$ from our calculations.

Q: What does it mean that the rod is massless?

A massless rod means we do not have to account for the gravitational force acting on the rod itself. If the rod had mass, we would have to include an additional downward force of $Mg$ acting at its center of mass ($l/2$), which would add another torque term to our equilibrium equation.

Q: If the left string has twice the frequency, why is its tension four times greater, not twice?

Because the fundamental frequency is proportional to the square root of tension ($f \propto \sqrt{T}$). Therefore, to double the frequency ($f_{AB} = 2 f_{CD}$), we must quadruple the tension ($T_{AB} = 2^2 T_{CD} = 4 T_{CD}$).

Q: Would the answer change if the strings had different lengths?

Yes. The fundamental frequency is inversely proportional to the length of the string ($f \propto 1/L$). If the strings had different lengths, that length ratio would enter the frequency equation, changing the required tension ratio and consequently shifting the position $x$.

Question

A massless rod $BD$ is suspended by two identical massless strings $AB$ and $CD$ of equal lengths. A block of mass $m$ is suspended from point $P$ such that $BP$ is equal to $x$. If the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of $x$ is

Options:

(A) $\frac{l}{5}$

(B) $\frac{l}{4}$

(C) $\frac{4l}{5}$

(D) $\frac{3l}{4}$

Sigma Percentile · Accepted Answer

### Introduction

Imagine you are standing in a physics lab, looking at a horizontal rod suspended by two vertical wires.

At first glance, this looks like a straightforward problem in **static equilibrium**—the kind of problem you solve in the first few weeks of mechanics.

But then, someone plucks the two vertical wires. They begin to hum, producing pure musical notes.

Suddenly, this simple mechanical system is transformed into a beautiful bridge between two seemingly disconnected worlds: **wave mechanics** and **rotational equilibrium**.

This is the essence of JEE Advanced physics. It takes two fundamental concepts, weaves them together, and asks you to find the elegant thread that connects them.

Let's embark on a journey to solve this problem step-by-step, understanding not just the *how*, but the deep physical *why* behind every equation.

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### Analyzing the Setup

Let's first visualize the physical reality of our system.

We have a horizontal rod $BD$ of length $l$. The problem states that the rod is **massless**. This is a crucial simplification! It means we do not have to worry about the weight of the rod itself acting at its center of mass.

The rod is suspended at its ends by two identical vertical strings, $AB$ on the left and $CD$ on the right. Because these strings are identical, they have the same length $L$ and the same linear mass density $\mu$.

Now, we hang a block of mass $m$ from a point $P$ on the rod, located at a distance $x$ from the left end $B$.

$$	ext{Distance } BP = x$$
$$	ext{Distance } PD = l - x$$

This hanging mass is the sole source of force acting downwards on our massless rod. The rod, in turn, pulls down on the two vertical strings, creating tensions $T_{AB}$ and $T_{CD}$ in them.

Our goal is to find the exact position $x$ where we must hang this mass so that the fundamental frequency of the left wire is exactly twice that of the right wire.

---

### The Wave Mechanics Connection

Let's start by analyzing the vertical strings. When a string of length $L$ and linear mass density $\mu$ is stretched under a tension $T$, transverse waves travel along it with a speed:

$$v = \sqrt{\frac{T}{\mu}}$$

When the string vibrates in its fundamental mode, a standing wave is formed with nodes at both clamped ends. The wavelength of this fundamental mode is twice the length of the string:

$$\lambda = 2L$$

Since wave speed is the product of frequency and wavelength ($v = f \lambda$), the fundamental frequency $f$ is:

$$f = \frac{v}{\lambda} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}$$

Look closely at this formula. Since both strings are identical, their length $L$ and linear mass density $\mu$ are constant.

This reveals a beautiful, simple proportionality:

$$f \propto \sqrt{T}$$

The fundamental frequency of vibration is directly proportional to the square root of the tension. A tighter string vibrates faster and produces a higher pitch!

---

### The Tension Ratio

We are given that the fundamental frequency of the left wire ($AB$) is twice that of the right wire ($CD$):

$$f_{AB} = 2 f_{CD}$$

Using our proportionality $f \propto \sqrt{T}$, we can substitute the square root of tension for frequency:

$$\sqrt{T_{AB}} = 2 \sqrt{T_{CD}}$$

To find the direct relationship between the tensions, we square both sides of this equation:

$$T_{AB} = 4 T_{CD}$$

This is our first major breakthrough! To make the left string vibrate at twice the frequency of the right string, the tension in the left string must be exactly **four times** the tension in the right string.

---

### The Art of Balance: Rotational Equilibrium

Now that we know how the tensions must be related, how do we physically achieve this tension distribution?

This is where **rotational mechanics** comes to the rescue.

Since the rod is in static equilibrium, it is not rotating. Therefore, the net torque acting on the rod about *any* pivot point must be exactly zero:

$$\Sigma 	au = 0$$

We can choose any point on the rod as our pivot. To make our algebra as clean as possible, let's choose the suspension point $P$ (where the mass $m$ is hung) as our pivot.

Why is this choice so clever?

Because the downward force of gravity $mg$ acts directly at point $P$. Since its distance from the pivot is zero, its torque is also zero:

$$	au_{mg} = mg \cdot 0 = 0$$

This completely eliminates the unknown mass $m$ and gravity $g$ from our torque equation!

Now, let's look at the torques produced by the tensions at the ends of the rod:

1. The tension $T_{AB}$ pulls upward at the left end $B$, which is at a distance $x$ to the left of our pivot $P$. This force tends to rotate the rod counter-clockwise about $P$. The magnitude of this counter-clockwise torque is:
   $$	au_{	ext{ccw}} = T_{AB} \cdot x$$

2. The tension $T_{CD}$ pulls upward at the right end $D$, which is at a distance $(l - x)$ to the right of our pivot $P$. This force tends to rotate the rod clockwise about $P$. The magnitude of this clockwise torque is:
   $$	au_{	ext{cw}} = T_{CD} \cdot (l - x)$$

For the rod to remain perfectly balanced, these two opposing torques must cancel each other out:

$$	au_{	ext{ccw}} = 	au_{	ext{cw}}$$

$$T_{AB} \cdot x = T_{CD} \cdot (l - x)$$

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### The Final Synthesis

We now have two elegant equations:

1. From wave mechanics: $T_{AB} = 4 T_{CD}$
2. From rotational mechanics: $T_{AB} \cdot x = T_{CD} \cdot (l - x)$

Let's substitute the tension relation into our torque equation:

$$4 T_{CD} \cdot x = T_{CD} \cdot (l - x)$$

Since the tension $T_{CD}$ is non-zero, we can divide both sides of the equation by $T_{CD}$ to cancel it out:

$$4x = l - x$$

Now, let's solve for $x$. Add $x$ to both sides to group all the $x$ terms together:

$$5x = l$$

Finally, divide by $5$:

$$x = \frac{l}{5}$$

There we have it! The mass must be suspended at a distance of exactly **one-fifth of the length of the rod** from the left end $B$.

This perfectly matches **Option (a)**.

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### Summary and Key Takeaways

This problem is a masterclass in physical reasoning. Let's look back at the beautiful logic path we followed:

* We connected **frequency** to **tension** using the physics of standing waves on a string ($f \propto \sqrt{T}$).
* We discovered that a frequency ratio of $2:1$ requires a tension ratio of $4:1$.
* We applied **rotational equilibrium** about the suspension point to relate the tensions to the position $x$.
* By combining these two concepts, the tensions canceled out, leaving us with a pure geometric ratio: $x = l/5$.

Next time you see a musical instrument or a suspended bridge, remember: the music it makes and the balance it keeps are two sides of the same beautiful physical coin!

Select Answer:

Visualized Solution

Visualizing the Physical System

Connecting Frequency and Tension

Applying the Frequency Condition

Finding the Tension Relation

Rotational Equilibrium of the Rod

Formulating the Torque Equation

Substituting the Tension Relation

Solving for $x$

The Way Forward

The Sigma Insight: Standing Waves in Strings and Organ Pipes

Introduction

Analyzing the Setup

The Wave Mechanics Connection

The Tension Ratio

The Art of Balance

The Final Synthesis

Summary and Key Takeaways

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