A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit le

Q: Why do we use $L_0$ instead of $2L_0$?

Because frequency is inversely proportional to length ($f \propto 1/L$). To get the *highest* frequency, we must use the *shortest* possible length, which is $L_0$.

Q: What does 'fundamental mode' mean?

It means the string is vibrating in its simplest possible pattern, forming exactly one loop (or antinode) between the two fixed ends. In this mode, the length of the string is half the wavelength ($L = \lambda/2$).

Q: Why is the tension $T_0$ the same for all strings?

The problem explicitly states 'If the tension in each string is $T_0$'. This is a given constraint, meaning the tuning pegs are adjusted to provide the exact same pulling force on all four strings.

Q: How did we get the factor of $1/\sqrt{2}$ for the second string?

The mass per unit length for the second string is $2\mu$. Since $\mu$ is in the denominator inside the square root, replacing it with $2\mu$ introduces a factor of $1/\sqrt{2}$ to the entire expression.

Q: What if the question asked for the lowest fundamental frequency?

If it asked for the lowest frequency, we would want to maximize the length. We would use $L = 2L_0$, which would halve all the frequencies we just calculated.

Question

A musical instrument is made using four different metal strings, 1, 2, 3 and 4 with mass per unit length $\mu$, $2\mu$, $3\mu$ and $4\mu$ respectively. The instrument is played by vibrating the strings by varying the free length in between the range $L_0$ and $2L_0$. It is found that in string-1 ($\mu$) at free length $L_0$ and tension $T_0$ the fundamental mode frequency is $f_0$.

List-I gives the above four strings while list-II the magnitude of some quantity. If the tension in each string is $T_0$, the correct match for the highest fundamental frequency in $f_0$ units will be,

List-I:

P. String-1 ($\mu$)

Q. String-2 ($2\mu$)

R. String-3 ($3\mu$)

S. String-4 ($4\mu$)

List-II:

1. $1$

2. $1/2$

3. $1/\sqrt{2}$

4. $1/\sqrt{3}$

5. $3/16$

6. $1/16$

Sigma Percentile · Accepted Answer

### The Symphony of Strings

Imagine you are holding a guitar. When you pluck a string, it vibrates and produces a sound. The pitch of that sound—its frequency—depends on three fundamental properties of the string: how long it is, how tightly it is stretched, and how thick or heavy it is.

In this beautiful problem from JEE Advanced, we are essentially tuning a mathematical musical instrument. We have four different strings, each with a different mass per unit length, and we are asked to find the highest possible fundamental frequency they can produce.

Let's dive into the physics of how these strings sing!

### The Master Equation

The fundamental frequency $f$ of a stretched string is governed by a classic wave equation:

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

Let's break down what this equation is telling us. The term $L$ represents the vibrating length of the string. Notice that $L$ is in the denominator, which means frequency is inversely proportional to length. A shorter string produces a higher pitch.

The term $T$ is the tension in the string. It is in the numerator inside the square root, meaning a tighter string produces a higher frequency. Finally, $\mu$ is the mass per unit length (linear mass density). A heavier, thicker string will vibrate more slowly, producing a lower frequency.

### Decoding the Problem Statement

The problem introduces a reference state using the first string. For String-1, the mass per unit length is $\mu$. When its vibrating length is set to $L_0$ and the tension is $T_0$, it produces a fundamental frequency $f_0$.

We can write this reference state mathematically as:

$$f_0 = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}}$$

This $f_0$ will serve as our measuring stick. We need to express the maximum frequencies of all the strings in terms of this $f_0$.

### The Quest for the Highest Frequency

The core trick of this question lies in a single word: **highest**. We are asked to find the highest fundamental frequency for each string.

The problem states that the free length $L$ can be varied in the range from $L_0$ to $2L_0$. Since frequency $f$ is inversely proportional to length $L$, how do we maximize $f$?

We must minimize $L$!

Therefore, to get the highest frequency, we must set the length to its minimum possible value:

$$L_{min} = L_0$$

The tension is kept constant at $T_0$ for all strings. Now, all we have to do is plug in the different values of $\mu$ for each string and see what happens.

### Calculating for Each String

Let's calculate the maximum frequency for each of the four strings systematically.

**String-1:**
The mass per unit length is $\mu_1 = \mu$. Using our minimum length $L_0$:

$$f_{1, max} = \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} = f_0$$

In units of $f_0$, this is exactly $1$.

**String-2:**
The mass per unit length is doubled, $\mu_2 = 2\mu$.

$$f_{2, max} = \frac{1}{2L_0} \sqrt{\frac{T_0}{2\mu}}$$

We can pull out the factor of $2$ from the denominator inside the square root:

$$f_{2, max} = \frac{1}{\sqrt{2}} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{f_0}{\sqrt{2}}$$

In units of $f_0$, this is $1/\sqrt{2}$.

**String-3:**
The mass per unit length is tripled, $\mu_3 = 3\mu$.

$$f_{3, max} = \frac{1}{2L_0} \sqrt{\frac{T_0}{3\mu}}$$

Pulling out the factor of $3$:

$$f_{3, max} = \frac{1}{\sqrt{3}} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{f_0}{\sqrt{3}}$$

In units of $f_0$, this is $1/\sqrt{3}$.

**String-4:**
The mass per unit length is quadrupled, $\mu_4 = 4\mu$.

$$f_{4, max} = \frac{1}{2L_0} \sqrt{\frac{T_0}{4\mu}}$$

The square root of $4$ is $2$, so we get:

$$f_{4, max} = \frac{1}{2} \left( \frac{1}{2L_0} \sqrt{\frac{T_0}{\mu}} \right) = \frac{f_0}{2}$$

In units of $f_0$, this is $1/2$.

### The Final Match

We have successfully calculated the highest fundamental frequencies for all four strings. Now, we simply match our results with the given lists:

*   **String-1 (I)** matches with **$1$ (P)**.
*   **String-2 (II)** matches with **$1/\sqrt{2}$ (R)**.
*   **String-3 (III)** matches with **$1/\sqrt{3}$ (S)**.
*   **String-4 (IV)** matches with **$1/2$ (Q)**.

The physics is elegant, the math is clean, and the symphony is complete!

List-I

List-II

Select Matching Pairs:

Visualized Solution

Fundamental Mode

Frequency Formula

Reference State $f_{0}$

Maximizing Frequency

String-1 Calculation

String-2 Calculation

String-3 Calculation

String-4 Calculation

Matrix Match

Conclusion

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Strings

The Master Equation

Decoding the Problem Statement

The Quest for the Highest Frequency

Calculating for Each String

The Final Match

Similar Questions

List-I

List-II

Two identical strings $X$ and $Z$ made of same material have tensions $T_{X}$ and $T_{Z}$ in them. If their fundamental frequencies are $450 Hz$ and $300 Hz$ respectively, then the ratio of $T_{X} / T_{Z}$ is

A wire of length $L$ and mass per unit length $6.0 \times 1 0^{- 3} kgm^{- 1}$ is put under tension of $540 N$ . Two consecutive frequencies that it resonates at are : $420 Hz$ and $490 Hz$ . Then, $L$ in metres is

A string of length $1 m$ and mass $2 \times 1 0^{- 5} kg$ is under tension $T$ . when the string vibrates, two successive harmonics are found to occur at frequencies $750 Hz$ and $1000 Hz$ . The value of tension $T$ is _______ Newton.

List-I

List-II

Select Matching Pairs:

Visualized Solution

Fundamental Mode

Frequency Formula

Reference State .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f0​

Maximizing Frequency

String-1 Calculation

String-2 Calculation

String-3 Calculation

String-4 Calculation

Matrix Match

Conclusion

The Sigma Insight: Standing Waves in Strings and Organ Pipes

The Symphony of Strings

The Master Equation

Decoding the Problem Statement

The Quest for the Highest Frequency

Calculating for Each String

The Final Match

Similar Questions

List-I

List-II

Reference State $f_{0}$