Problem 42 - University Physics with Modern Physics

Question

A piano tuner stretches a steel piano wire with a tension of 800 \(\mathrm{N}\) . The steel wire is 0.400 \(\mathrm{m}\) long and has a mass of 3.00 \(\mathrm{g}\) . (a) What is the frequency of its fundamental mode of vibration? (b) What is the number of the highest harmonic that could be heard by a person who is capable of hearing frequencies up to \(10,000 \mathrm{Hz} ?\)

Step-by-Step Solution

Answer

(a) The fundamental frequency is 471.4 Hz. (b) The highest audible harmonic is the 21st.

Step 1: Convert Mass to Kilograms

The mass of the wire is given in grams. Convert it to kilograms by dividing by 1000, since there are 1000 grams in a kilogram:\[ 3.00 \, \text{g} = \frac{3.00}{1000} \, \text{kg} = 0.003 \, \text{kg} \]

Step 2: Calculate Linear Mass Density

The linear mass density \( \mu \) is the mass per unit length of the wire. Calculate it by dividing the mass by the length:\[ \mu = \frac{m}{L} = \frac{0.003 \, \text{kg}}{0.400 \, \text{m}} = 0.0075 \, \text{kg/m} \]

Step 3: Use the Formula for Fundamental Frequency

The fundamental frequency of a stretched string is given by \( f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( T \) is the tension, \( L \) is the length, and \( \mu \) is the linear mass density. Plug in the values:\[ f_1 = \frac{1}{2 \times 0.400 \, \text{m}} \sqrt{\frac{800 \, \text{N}}{0.0075 \, \text{kg/m}}} \]Calculate \( f_1 \):\[ f_1 = 471.4 \, \text{Hz} \]

Step 4: Determine the Number of Highest Audible Harmonic

A person can hear frequencies up to 10,000 Hz. Use the fundamental frequency to find the highest harmonic by calculating \( n \) such that \( n \times f_1 \leq 10,000 \) Hz. Solve for \( n \):\[ n \leq \frac{10,000}{471.4} \approx 21.21 \]Since \( n \) must be an integer, the highest audible harmonic number is 21.

Key Concepts

Fundamental Frequency

When we talk about the fundamental frequency, we're discussing the lowest frequency at which a system, like a stretched wire, naturally vibrates. In the context of musical instruments, this is the pitch you hear when a single note is played. For a string fixed at both ends, the fundamental frequency depends on several important factors:

• Length of the string (L)
• The tension in the string (T)
• Linear mass density of the string (\( \mu \))

The formula to calculate this frequency is \[ f_1 = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]This formula illustrates how shortening the string, increasing tension, or reducing mass will increase the fundamental frequency. For practical contexts like a piano, tuning is achieved by adjusting tension, to match specific pitches. By substituting the given values into the formula, for example with a steel wire of length 0.400 m, tension 800 N, and linear mass density 0.0075 kg/m, we calculated a fundamental frequency of approximately 471.4 Hz.

Harmonics

Harmonics are fascinating extensions of the fundamental frequency. They are generated when a string vibrates at whole-number multiples of the fundamental frequency. If the fundamental frequency is \( f_1 \), the second harmonic would be \( 2f_1 \), the third \( 3f_1 \), and so on. These harmonics contribute to the richness of sound, providing overtones that shape the character of musical notes.
In the case of our piano wire, we need to find out how many harmonics are audible to a typical human ear, which hears from around 20 Hz up to 10,000 Hz. By dividing the upper frequency limit by the fundamental frequency, \( 10,000 \div 471.4 \approx 21.21 \), we discover that up to the 21st harmonic fits within the hearing range. Thus, the highest harmonic number that can be perceived is 21. Each harmonic contributes to the sound's timbre but may not always be individually distinguished.

Linear Mass Density

Linear mass density \( \mu \) is a key concept in understanding how strings and wires vibrate. Imagine this as the distribution of mass along the string's length. In a formal sense, it's defined as the mass per unit length:\[ \mu = \frac{m}{L} \]where \( m \) is the mass and \( L \) is the length of the string.
For the piano wire in question, with a total mass of 3.00 g and a length of 0.400 m, you convert the mass to kilograms (0.003 kg) before calculating the density. This gives us a linear mass density of 0.0075 kg/m. This variable is important because it provides resistance to the linear motion, influencing how fast waves (and therefore frequencies) travel through the string. A higher \( \mu \) will result in a lower fundamental frequency, while a lower \( \mu \) allows for a higher pitch. Using this concept, tuners can adjust strings by changing tension without altering the fundamental physical characteristics of the string.