Questions: A trumpet plays its 3rd harmonic at 510 Hz. It then opens a valve, which adds 0.110 m to its length. What is the new 3rd harmonic frequency? (Hint: Find the original length.) (Speed of sound = 343 m / s ) (Unit = Hz )

Solution Steps

The 3rd harmonic frequency of an open pipe is given by: \[ f_3 = \frac{3v}{2L} \] where \( f_3 \) is the 3rd harmonic frequency, \( v \) is the speed of sound, and \( L \) is the length of the pipe.

Given: \[ f_3 = 510 \, \text{Hz} \] \[ v = 343 \, \text{m/s} \]

Rearranging the formula to solve for \( L \): \[ L = \frac{3v}{2f_3} \]

Substituting the given values: \[ L = \frac{3 \times 343}{2 \times 510} \] \[ L = \frac{1029}{1020} \] \[ L \approx 1.0088 \, \text{m} \]

When the valve is opened, the length of the trumpet increases by 0.110 m: \[ L_{\text{new}} = L + 0.110 \] \[ L_{\text{new}} = 1.0088 + 0.110 \] \[ L_{\text{new}} = 1.1188 \, \text{m} \]

Using the new length \( L_{\text{new}} \) to find the new 3rd harmonic frequency: \[ f_{3,\text{new}} = \frac{3v}{2L_{\text{new}}} \]

Substituting the new length: \[ f_{3,\text{new}} = \frac{3 \times 343}{2 \times 1.1188} \] \[ f_{3,\text{new}} = \frac{1029}{2.2376} \] \[ f_{3,\text{new}} \approx 459.8 \, \text{Hz} \]

Final Answer