JEE Advanced 2000

LEVELJEE Main

Animated Solution for Physics - Waves: Two vibrating strings of the same material but of lengths $L$ and $2 L$ have radii $2 r$ and $r$ respectively. They are stretched under the same tension. Both the strings vibrate in their fundamental modes. The one of length $L$ with frequency $ν_{1}$ and the other with frequency $ν_{2}$ . The ratio $ν_{1} / ν_{2}$ is given by

Q: Why does the mass per unit length depend on the radius squared?

Because strings are cylindrical. The mass per unit length $\mu$ is the mass of a 1-meter segment of the string. Since mass is density times volume, and the volume of a 1-meter cylinder is its cross-sectional area times length ($A \times 1 = \pi r^2$), $\mu$ is directly proportional to the cross-sectional area, which depends on $r^2$.

Q: Does the material of the string affect the frequency?

Yes, absolutely! The material determines the density $\rho$. A denser material (like steel vs. nylon) will have a larger mass per unit length $\mu$, which decreases the wave speed and thus lowers the fundamental frequency. In this problem, because both strings are of the 'same material', the density $\rho$ is identical and cancels out.

Q: What would happen to the ratio if the strings were vibrating in different harmonics?

If they were in different harmonics, say the $n_1$-th and $n_2$-th harmonics, the frequency formula would include the harmonic number: $f = \frac{n}{2l}\sqrt{\frac{T}{\mu}}$. The ratio would then be $\frac{\nu_1}{\nu_2} = \frac{n_1}{n_2} \left(\frac{l_2 r_2}{l_1 r_1}\right)$.

Q: Is the wave speed the same in both strings?

No, the wave speed is different! Since $v = \sqrt{T/\mu}$ and the thicker string has a larger $\mu$, the wave speed is slower in the thicker string. Specifically, the wave speed in the first string is half of that in the second string ($v_1 = v_2 / 2$).

Select Answer:

Visualized Solution

Visualizing the Two Vibrating Strings

We are given two strings of the same material under the same tension $T$ .
String 1 has length $L$ and radius $2 r$ , vibrating with frequency $ν_{1}$ .
String 2 has length $2 L$ and radius $r$ , vibrating with frequency $ν_{2}$ .
Both strings vibrate in their fundamental modes.

The Fundamental Frequency Formula

The fundamental frequency $f$ of a stretched string of length $l$ fixed at both ends is given by:
$f = \frac{v}{2 l} = \frac{1}{2 l} \frac{T}{μ}$
where $T$ is the tension and $μ$ is the mass per unit length of the string.

Relating Mass per Unit Length to Radius

The mass per unit length $μ$ can be expressed in terms of the material's density $ρ$ and cross-sectional area $A$ :
$μ = \frac{mass}{length} = \frac{ρ \cdot V}{l} = \frac{ρ \cdot A \cdot l}{l} = ρ A$
Since the cross-section is circular with radius $r$ :
$μ = ρ (π r^{2})$

Proportionality of Frequency

Substituting $μ = ρ π r^{2}$ into the frequency formula:
$f = \frac{1}{2 l} \frac{T}{ρ π r ^{2}} = \frac{1}{2 l r} \frac{T}{ρ π}$
Since the material (density $ρ$ ) and tension $T$ are the same for both strings:
$f \propto \frac{1}{l \cdot r}$

Setting up the Ratio

Using the proportionality $f \propto \frac{1}{l \cdot r}$ , we can write the ratio of frequencies as:
$\frac{ν _{1}}{ν _{2}} = \frac{l _{2} \cdot r _{2}}{l _{1} \cdot r _{1}}$

Substituting the Given Values

We are given:
- For String 1: $l_{1} = L$ , $r_{1} = 2 r$
- For String 2: $l_{2} = 2 L$ , $r_{2} = r$
Substituting these values into the ratio:
$\frac{ν _{1}}{ν _{2}} = \frac{( 2 L ) \cdot ( r )}{( L ) \cdot ( 2 r )}$

Calculating the Final Ratio

Simplifying the expression:
$\frac{ν _{1}}{ν _{2}} = \frac{2 L r}{2 L r} = 1$
Thus, the ratio of frequencies is $1$ , which corresponds to option (d).

Exploring Further Variations

What if the tensions were different? Or if the materials were different?
If $T_{1} \neq = T_{2}$ and $ρ_{1} \neq = ρ_{2}$ , then:
$\frac{ν _{1}}{ν _{2}} = \frac{l _{2} r _{2}}{l _{1} r _{1}} \frac{T _{1} ρ _{2}}{T _{2} ρ _{1}}$
This shows how JEE can twist this simple question by changing other physical parameters.

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The Sigma Insight: Standing Waves in Strings and Organ Pipes

Solution Diagram

Analyzing the Setup

Imagine you are holding two different guitar strings made of the exact same material and stretched under the same tension

T

One string is short and thick, while the other is long and thin.

Intuitively, you might expect them to produce completely different notes. But physics has a beautiful way of balancing things out. Let's dive deep into the mathematics of standing waves to see how these geometric differences interact.

The Master Equation

For any string of length

l

fixed at both ends, the fundamental mode of vibration consists of a single loop with nodes at the ends and an antinode in the middle.

The wavelength

λ

of this fundamental mode is twice the length of the string:

λ = 2 l

The fundamental frequency

f

is related to the wave speed

v

by:

f = \frac{v}{λ} = \frac{v}{2 l}

Recall that the speed of a transverse wave on a stretched string depends on the tension

T

and the mass per unit length

μ

v = \frac{T}{μ}

Substituting this back into our frequency equation gives us our master tool:

f = \frac{1}{2 l} \frac{T}{μ}

Unveiling the Role of Radius

To understand how the thickness (radius

r

) of the string affects the frequency, we must express the mass per unit length

μ

in terms of the string's geometry and material density

ρ

By definition,

μ

is the mass of a unit length of the string:

μ = \frac{Mass}{Length} = \frac{ρ \cdot Volume}{l}

Since a string is a cylinder of cross-sectional area

A = π r^{2}

and length

l

, its volume is

V = A \cdot l = π r^{2} l

. Substituting this in:

μ = \frac{ρ ( π r ^{2} l )}{l} = ρ π r^{2}

Now, let's substitute this expression for

μ

back into our master frequency equation:

f = \frac{1}{2 l} \frac{T}{ρ π r ^{2}} = \frac{1}{2 l r} \frac{T}{ρ π}

The Power of Proportionality

Since both strings are made of the same material (same density

ρ

) and are stretched under the same tension

T

, the term inside the square root is a constant for both strings:

\frac{T}{ρ π} = Constant

Therefore, the fundamental frequency is inversely proportional to the product of the length

l

and the radius

r

f \propto \frac{1}{l \cdot r}

This is a remarkably elegant result! It tells us that a change in length can be exactly compensated by a corresponding change in radius.

Final Calculation

Let's set up the ratio of the frequencies of the two strings:

\frac{u _{1}}{u _{2}} = \frac{l _{2} \cdot r _{2}}{l _{1} \cdot r _{1}}

We are given the following parameters: - For String 1:

l_{1} = L

r_{1} = 2 r

- For String 2:

l_{2} = 2 L

r_{2} = r

Substituting these values into our ratio:

\frac{u _{1}}{u _{2}} = \frac{( 2 L ) \cdot ( r )}{( L ) \cdot ( 2 r )}

\frac{u _{1}}{u _{2}} = \frac{2 L r}{2 L r} = 1

Despite their different physical dimensions, the two strings vibrate at the exact same fundamental frequency!

The geometric changes perfectly cancel each other out, yielding a ratio of 1, which corresponds to option (d).

Common Traps & FAQs

Why does the mass per unit length depend on the radius squared?

Because strings are cylindrical. The mass per unit length

μ

is the mass of a 1-meter segment of the string. Since mass is density times volume, and the volume of a 1-meter cylinder is its cross-sectional area times length (

A \times 1 = π r^{2}

μ

is directly proportional to the cross-sectional area, which depends on

r^{2}

Does the material of the string affect the frequency?

Yes, absolutely! The material determines the density

ρ

. A denser material (like steel vs. nylon) will have a larger mass per unit length

μ

, which decreases the wave speed and thus lowers the fundamental frequency. In this problem, because both strings are of the 'same material', the density

ρ

is identical and cancels out.

What would happen to the ratio if the strings were vibrating in different harmonics?

If they were in different harmonics, say the

n_{1}

-th and

n_{2}

-th harmonics, the frequency formula would include the harmonic number:

f = \frac{n}{2 l} \frac{T}{μ}

. The ratio would then be

\frac{ν _{1}}{ν _{2}} = \frac{n _{1}}{n _{2}} (\frac{l _{2} r _{2}}{l _{1} r _{1}})

Why does the tension remain the same if one string is thicker?

Tension is an external pulling force applied to stretch the string. It is independent of the string's thickness. The problem explicitly states that both strings are stretched under the same tension, meaning the external stretching force is kept constant for both.

Is the wave speed the same in both strings?

No, the wave speed is different! Since