Explanation

Wave speed in a stretched string is:

\[ v = \sqrt{\frac{T}{\mu}} \]

Since tension is same (500 N) in both strings, wave speeds differ due to different linear mass densities.

For string A:

\[ v_A = \sqrt{\frac{500}{2 \times 10^{-4}}} \]

For string B:

\[ v_B = \sqrt{\frac{500}{4 \times 10^{-4}}} \]

Time taken by pulse to reach junction:

\[ t_1 = \frac{L_A}{v_A} \]

\[ t_2 = \frac{L_B}{v_B} \]

Taking ratio:

\[ \frac{t_1}{t_2} = \frac{L_A}{L_B} \sqrt{\frac{\mu_A}{\mu_B}} \]

\[ = \frac{2.5}{1.5} \sqrt{\frac{2}{4}} \]

\[ = \frac{5}{3} \times \frac{1}{\sqrt{2}} \]

\[ \approx 1.18 \]

Therefore correct answer is (B).

Related Theory (Complete Mechanical Waves Mastery Section)

This question is based on transverse wave propagation in stretched strings, one of the most fundamental topics in mechanical waves. Understanding wave speed, linear mass density, and tension relationships is essential for JEE Main and JEE Advanced.

1. Derivation of Wave Speed Formula

Consider a small element of string under tension T. When a transverse disturbance passes, restoring force arises due to tension. Applying Newton’s second law to the element leads to the wave equation:

\[ \frac{\partial^2 y}{\partial t^2} = \frac{T}{\mu} \frac{\partial^2 y}{\partial x^2} \]

Comparing with standard wave equation:

\[ \frac{\partial^2 y}{\partial t^2} = v^2 \frac{\partial^2 y}{\partial x^2} \]

We get:

\[ v = \sqrt{\frac{T}{\mu}} \]

This formula shows wave speed depends only on medium properties (T and μ), not on amplitude or frequency.

2. Physical Interpretation

If tension increases → string becomes stiffer → wave travels faster.

If linear density increases → string becomes heavier → wave travels slower.

3. Mechanical Impedance

Mechanical impedance of a string:

\[ Z = \sqrt{T\mu} \]

When pulse reaches junction of two strings with different μ, impedance mismatch causes reflection and transmission.

If wave travels from lighter to heavier string:

• Reflected pulse is inverted.

If wave travels from heavier to lighter string:

• Reflected pulse is not inverted.

Though reflection behavior is not required here, it is a frequently tested JEE concept.

4. Time of Travel Concept

Time taken for pulse to travel a distance L:

\[ t = \frac{L}{v} \]

Substituting v:

\[ t = L \sqrt{\frac{\mu}{T}} \]

Thus time is directly proportional to length and square root of linear density.

5. Why Tension is Same?

Since both strings are connected and stretched between rigid supports, equilibrium ensures same tension throughout.

This is a common conceptual trap. Students sometimes assume tension differs due to different μ.

6. Advanced Concept: Energy Flow

Power transmitted by wave:

\[ P = \frac{1}{2}\mu \omega^2 A^2 v \]

Energy transport depends on both μ and v.

7. Common Mistakes in Exams

• Forgetting square root relation.
• Taking μ ratio instead of √μ ratio.
• Ignoring length ratio.
• Using incorrect units.

8. JEE Pattern Insight

Wave speed formula is directly asked in many forms:

• Change in tension → change in frequency
• Two strings joined → reflection behavior
• Standing waves in composite strings
• Ratio of times or frequencies

9. Conceptual Summary

• Wave speed depends only on T and μ.
• Heavier string → slower wave.
• Time = length / speed.
• For same tension → ratio depends on √μ and length.

Mechanical waves form strong scoring unit in JEE Main.

Mastering this relationship ensures quick and accurate solving of wave motion problems.

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