Here's a breakdown of how to solve this problem:
**Understanding Harmonics in Open-Closed Tubes**
* A clarinet can be approximated as a tube open at one end (mouthpiece) and closed at the other (the first open tone hole).
* In such tubes, only odd harmonics are produced. This means the frequencies are odd-numbered multiples of the fundamental frequency.
**Calculations**
1. **Harmonic Calculation:**
 * The fundamental frequency (D3) is 146.83 Hz.
 * A4 has a frequency of 440 Hz.
 * Divide the frequency of A4 by the fundamental frequency: 440 Hz / 146.83 Hz ≈ 3
 * **The clarinet is operating at the 3rd harmonic.**
2. Â **Length Calculation**
  * **Speed of Sound:** The speed of sound in air at 20 degrees C is approximately 343 m/s.
  * **Wavelength:**
    * For the 3rd harmonic in an open-closed tube, the wavelength is four times the tube's length (λ = 4L).
    * Wavelength can be calculated as speed/frequency (λ = v/f)
    * λ = 343 m/s / 440 Hz ≈ 0.78 m
  * **Tube Length:**
    * Since λ = 4L, then L = λ / 4
    * L = 0.78 m / 4 ≈ 0.195 m
**Therefore:**
* The clarinet is operating at the **3rd harmonic** when playing A4.
* The length of the clarinet is approximately **0.195 meters** (or 19.5 cm).
