Physics · Sound Waves

JEE Main 2024 — 6 April, Shift 2 — Question 59

Two open organ pipes of length 60 cm and 90 cm resonate at $6^{\text {th }}$ and $5^{\text {th }}$ harmonics respectively. The difference of frequencies for the given modes is $\qquad$ Hz. (Velocity of sound in air $=333 \mathrm{~m} / \mathrm{s}$ )

Answer: 740

Numerical answer — enter this value.

Step-by-step solution

The difference in frequency in open organ pipe $=$

$\mathrm{f}=\frac{\mathrm{nV}}{2 \mathrm{~L}}$ $\Delta \mathrm{f}=\frac{6 \mathrm{v}}{2 \times 0.6}-\frac{5 \mathrm{v}}{2 \times 0.9}$

$\mathrm{v}=333 \mathrm{~m} / \mathrm{s}$

$\Delta \mathrm{f}=740 \mathrm{~Hz}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 2
Subject: Physics
Chapter: Sound Waves
Topic: Vibrations in rod and Air Columns - Organ pipes