QJEEM24Ph0e96: JEE Main 2024 Physics - Class 11 JEEM24Ph0e96 Solution: Two open organ pipes of lengths and resonate at and harmonics…

To solve this problem, let's first understand how the harmonics of open organ pipes work. For an open organ pipe, the harmonics are given by the formula:

$f_n = n \frac{v}{2L}$

where:

• $f_n$ is the frequency of the nth harmonic,


• $n$ is the harmonic number (an integer),


• $v$ is the speed of sound in air,


• $L$ is the length of the organ pipe.

Given that the speeds of sound in air $v = 333 \, \text{m/s}$ , and the lengths of the two open organ pipes are $60 \, \text{cm} = 0.60 \, \text{m}$ and $90 \, \text{cm} = 0.90 \, \text{m}$ , we can calculate the frequencies of the 6th harmonic for the 60 cm pipe and the 5th harmonic for the 90 cm pipe.

For the 60 cm pipe at the 6th harmonic ( $n = 6$ ):

$f_{6,60} = 6 \frac{333}{2 \times 0.60} = 6 \times \frac{333}{1.2} = 6 \times 277.5 = 1665 \, \text{Hz}$

For the 90 cm pipe at the 5th harmonic ( $n = 5$ ):

$f_{5,90} = 5 \frac{333}{2 \times 0.90} = 5 \times \frac{333}{1.8} = 5 \times 185 = 925 \, \text{Hz}$

The difference in frequencies between these two modes is:

$\Delta f = f_{6,60} - f_{5,90} = 1665 \, \text{Hz} - 925 \, \text{Hz} = 740 \, \text{Hz}$

Therefore, the difference of frequencies for the given modes is $740 \, \text{Hz}$ .