For a certain organ pipe, the first three resonance frequencies are in the ratio of $1:3:5$ respectively. If the frequency of fifth harmonic is $405 \mathrm{~Hz}$ and the speed of sound in air is $324 \mathrm{~ms}^{-1}$ the length of the organ pipe is _________ $\mathrm{m}$.

<p>Given that the first three resonance frequencies are in the ratio of $1:3:5$, we can express them as follows:</p> <p>$f_1 = kf$ $f_3 = 3kf$ $f_5 = 5kf$</p> <p>Where $k$ is a constant and $f_1, f_3$, and $f_5$ are the first, third, and fifth resonance frequencies, respectively. We are given that the frequency of the fifth harmonic is $405 \mathrm{~Hz}$, so we can write:</p> <p>$f_5 = 5kf = 405 \mathrm{~Hz}$</p> <p>Now we can solve for the constant $k$:</p> <p>$k = \frac{405}{5} = 81 \mathrm{~Hz}$</p> <p>We also know that the speed of sound in air is $v = 324 \mathrm{~ms}^{-1}$. The relationship between the speed of sound, the frequency, and the wavelength of a standing wave in a closed pipe can be expressed as follows:</p> <p>$v = f\lambda$</p> <p>Where $\lambda$ is the wavelength of the wave. For the first harmonic in a closed pipe, the length of the pipe is equal to one-fourth of the wavelength:</p> <p>$L = \frac{1}{4}\lambda$</p> <p>We can now substitute the expression for the wavelength in terms of the length into the equation for the speed of sound:</p> <p>$v = f_1 \cdot 4L$</p> <p>Now, we can substitute the value of $f_1 = kf = 81 \mathrm{~Hz}$ and the speed of sound $v = 324 \mathrm{~ms}^{-1}$ into the equation:</p> <p>$324 = 81 \times 4L$</p> <p>Now we can solve for the length of the organ pipe $L$:</p> <p>$L = \frac{324}{81 \times 4} = \frac{324}{324} = 1 \mathrm{~m}$</p> <p>The length of the organ pipe is $1 \mathrm{~m}$.</p>