An organ pipe is open at both ends. The frequency of the third mode is 320 Hz higher than the frequency of the second mode. If the speed of sound is 345 m/s, then what is the length of the organ pipe?

Sure, let's solve this step by step.

1. First, we need to understand the modes of vibration in an organ pipe. For a pipe open at both ends, the fundamental frequency (or first mode) occurs when the length of the pipe is half the wavelength of the sound. The second mode occurs when the length of the pipe is equal to the wavelength, and the third mode occurs when the length of the pipe is 1.5 times the wavelength.

2. The problem states that the frequency of the third mode is 320 Hz higher than the frequency of the second mode. In terms of wavelengths, this means that the third mode has 1.5 wavelengths fitting into the pipe, while the second mode has 1 wavelength fitting into the pipe. Therefore, the difference in frequency corresponds to 0.5 wavelengths.

3. We can use the formula for the speed of sound, v = fλ, where v is the speed of sound, f is the frequency, and λ is the wavelength. We can rearrange this formula to find the wavelength: λ = v/f.

4. We know that the speed of sound is 345 m/s and the difference in frequency is 320 Hz, so we can substitute these values into the formula to find the wavelength difference: λ = 345 m/s / 320 Hz = 1.078125 m.

5. Since this wavelength difference corresponds to 0.5 wavelengths in the pipe, we can multiply by 2 to find the total length of the pipe: L = 2 * 1.078125 m = 2.15625 m.

So, the length of the organ pipe is approximately 2.16 meters.