JEE Mains 2025 | Transverse Wave Speed Problem | 8th April Shift 2 Solved
If you’re preparing for JEE Mains 2025, understanding problems related to Transverse Waves is crucial. In this post, we’ll solve a real question from the 8th April Shift 2 Physics paper that tests your concept of wave speed in strings of different radii. This guide will help you master the method and avoid mistakes in the exam.
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This video explains the complete step-by-step solution. Watch it first before reading the breakdown below.
Question
Given:
- Two strings made of the same material and kept under the same tension.
- The first string has radius R.
- The second string has radius R/2.
- v1 and v2 are the transverse wave speeds in the first and second strings, respectively.
Find: The ratio v2 / v1
Options:
- 1/2
- 2
- 1
- 1/4
Detailed Step-by-Step Solution
Step 1: Use the Formula for Transverse Wave Speed
The transverse wave speed in a string is given by:
v = √(T / μ)
Where:
- T is the tension (same for both strings)
- μ is the linear mass density
Step 2: Calculate the Linear Mass Density
Linear mass density μ is proportional to the cross-sectional area of the string:
μ = ρ × A = ρ × πr²
- For the first string: μ1 = ρ × π × R²
- For the second string: μ2 = ρ × π × (R/2)² = ρ × π × R²/4
Step 3: Find the Ratio of Speeds
Using the wave speed formula:
v2 / v1 = √(μ1 / μ2) = √((ρ × π × R²) / (ρ × π × R²/4)) = √(4) = 2
Step 4: Choose the Correct Answer
The correct option is (2).
Conclusion
In this question, we learned how the transverse wave speed in a string depends on its radius. By applying the formula correctly, we found that v2 / v1 = 2.