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

<p>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: <p>$f_n = n \frac{v}{2L}$</p> <p>where:</p> <ul> <li>$f_n$ is the frequency of the nth harmonic,</li><br> <li>$n$ is the harmonic number (an integer),</li><br> <li>$v$ is the speed of sound in air,</li><br> <li>$L$ is the length of the organ pipe.</li> </ul> <p>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.</p> <p>For the 60 cm pipe at the 6th harmonic ($n = 6$):</p> <p>$$f_{6,60} = 6 \frac{333}{2 \times 0.60} = 6 \times \frac{333}{1.2} = 6 \times 277.5 = 1665 \, \text{Hz}$$</p> <p>For the 90 cm pipe at the 5th harmonic ($n = 5$):</p> <p>$$f_{5,90} = 5 \frac{333}{2 \times 0.90} = 5 \times \frac{333}{1.8} = 5 \times 185 = 925 \, \text{Hz}$$</p> <p>The difference in frequencies between these two modes is:</p> <p>$$\Delta f = f_{6,60} - f_{5,90} = 1665 \, \text{Hz} - 925 \, \text{Hz} = 740 \, \text{Hz}$$</p> <p>Therefore, the difference of frequencies for the given modes is $740 \, \text{Hz}$.</p></p>