Problem 67 - College Physics

Question

\(\bullet\) A person is playing a small flute 10.75 \(\mathrm{cm}\) long, open at one end and closed at the other, near a taut string having a fundamental frequency of 600.0 \(\mathrm{Hz}\) . If the speed of sound is \(344.0 \mathrm{m} / \mathrm{s},\) for which harmonics of the flute will the string res- onate? In each case, which harmonic of the string is in resonance?

Step-by-Step Solution

Answer

No harmonics of the flute naturally resonate with the string at 600 Hz.

Step 1: Understand the Problem

The problem involves finding the harmonics of a flute that will resonate with a string. The flute is open at one end and closed at the other, which affects how sound waves form inside it. We are given the fundamental frequency of the string (600 Hz) and the speed of sound (344 m/s).

Step 2: Determine the Resonating Condition

For resonance to occur, the frequency of a harmonic of the flute should match the frequency of a harmonic of the string. Therefore, we need to find the harmonics of the flute and check which ones are at or near 600 Hz.

Step 3: Frequency of Harmonics in the Flute

For a tube open at one end and closed at the other, the frequency of the nth harmonic is given by \(f_{n} = \frac{(2n-1) \cdot v}{4L}\). This applies to odd harmonics (e.g., 1st, 3rd, 5th, etc.). Here, \(L = 0.1075\) m and \(v = 344\) m/s.

Step 4: Calculate 1st Harmonic Frequency of the Flute

For n=1 (1st harmonic), \(f_{1} = \frac{(2*1-1) \cdot 344}{4 \times 0.1075} = \frac{344}{0.43} \approx 800\) Hz. This frequency doesn't match the string's 600 Hz.

Step 5: Calculate 3rd Harmonic Frequency of the Flute

For n=2 (3rd harmonic since only odd harmonics are present), \(f_{3} = \frac{(2*2-1) \cdot 344}{4 \times 0.1075} = \frac{3 \cdot 344}{0.43} \approx 2400\) Hz. This doesn't match 600 Hz either.

Step 6: Analyze Possible Resonance

Since none of the calculated flute harmonics match 600 Hz, no direct resonance occurs with the string's fundamental frequency. Therefore, we check if a lower harmonic of the string can match a higher harmonic of the flute.

Step 7: Determine the Correct Harmonics

For the string, if higher harmonics (like the 2nd harmonic) are considered, \(f_{n} = n \times 600\) Hz. If n=2, \(f_{2} = 1200\) Hz, which approximates conditions if errors are in the flute's computed frequencies. However, initially computed higher harmonics from the flute would logically not align unless instrument adjustments (e.g., temperature changes) occur.

Step 8: Conclusion of Harmonics

Given the problem constraints, the flute and the string are unlikely to naturally resonate without additional physical adjustments or factors unconsidered here; it suggests close checks of concept taught aligning practice variables.

Key Concepts

Harmonics

In the world of acoustics, harmonics refer to the multiple frequencies at which an object can naturally vibrate. For musical instruments like the flute, these vibrations create different pitches. A flute that is open at one end produces only odd harmonics, meaning these are the 1st, 3rd, 5th, and so on. In our problem, the flute's harmonics depend on its length and the speed of sound within it. By using the formula \( f_n = \frac{(2n-1) \cdot v}{4L} \), where \( f_n \) is the frequency of the nth harmonic, \( v \) is the speed of sound, and \( L \) is the length of the flute, we can determine the specific frequencies of the harmonics. For example, the 1st harmonic corresponds to the flute's fundamental frequency, and higher odd-numbered harmonics (like the 3rd) will produce higher pitches.

Resonance

Resonance happens when two systems, such as a flute and a string, vibrate at the same frequency. This phenomenon amplifies sound, creating a more robust tone. For resonance to occur in our problem, a harmonic from the flute must match the frequency of a harmonic from the string. The situation describes the flute and string potentially reaching resonance if any flute harmonic aligns closely with a string harmonic frequency. However, due to mismatches in initial harmonic calculations, the two instruments do not naturally resonate unless external conditions change or a different harmonic from the string is considered, enhancing the chance of resonance.

Fundamental Frequency

The fundamental frequency is the lowest frequency at which a system naturally vibrates. For a string, it's often the primary pitch heard, and is calculated based on length, tension, and mass per unit length. In this problem, the fundamental frequency of the string is given as 600 Hz. For the flute, which is open at one end and closed at the other, the fundamental frequency corresponds to its 1st harmonic, found using the \( f_n = \frac{(2n-1) \cdot v}{4L} \) formula. This primary frequency sets the stage for examining other possible harmonics, as it helps calculate multiples of higher related pitches.

Speed of Sound

The speed of sound, a crucial parameter in acoustics, influences how quickly sound waves travel through a medium. For air, this speed is approximately 344 m/s at room temperature. Sound speed impacts the calculation of frequencies for harmonics. For instance, in the flute exercise, it's a key component in determining the frequency of different harmonics using the formula \( f_n = \frac{(2n-1) \cdot v}{4L} \). The speed of sound can vary with temperature, pressure, and humidity changes, which can affect the perceived pitch of musical instruments, potentially modifying the conditions enough to achieve resonance where it wasn't initially observed.