Problem 49 - University Physics with Modern Physics

Question

A guitar string with a mass of \(10.0 \mathrm{~g}\) is \(1.00 \mathrm{~m}\) long and attached to the guitar at two points separated by \(65.0 \mathrm{~cm} .\) a) What is the frequency of the first harmonic of this string when it is placed under a tension of \(81.0 \mathrm{~N} ?\) b) If the guitar string is replaced by a heavier one that has a mass of \(16.0 \mathrm{~g}\) and is \(1.00 \mathrm{~m}\) long, what is the frequency of the replacement string's first harmonic?

Step-by-Step Solution

Answer

a) For the first string with a mass of 10.0 g: The frequency of the first harmonic is approximately 61.61 Hz. b) For the second string with a mass of 16.0 g: The frequency of the first harmonic is approximately 48.60 Hz.

Step 1: Calculate the linear mass density for the first string

Given the mass of the first string as \(10.0 \mathrm{~g}\) and the length as \(1.00 \mathrm{~m}\), the linear mass density can be calculated as follows: $$\mu_1 = \frac{10.0 \mathrm{~g}}{1.00 \mathrm{~m}} = 10.0 \mathrm{\frac gm}$$

Step 2: Calculate the first harmonic frequency for the first string

Using the formula for the fundamental frequency and the given tension of \(81.0 \mathrm{~N}\), the vibrating length of the first string as \(0.65 \mathrm{~m}\), and the calculated linear mass density: $$f_1 = \frac{1}{2(0.65 \mathrm{~m})} \sqrt{\frac{81.0 \mathrm{~N}}{10.0 \mathrm{\frac gm}}}$$ Converting grams to kilograms (to keep SI units): $$f_1 = \frac{1}{2(0.65 \mathrm{~m})} \sqrt{\frac{81.0 \mathrm{~N}}{0.01 \mathrm{\frac kgm}}} = 61.61 \mathrm{Hz}$$ So, the frequency of the first harmonic for the first string is approximately \(61.61 \mathrm{Hz}\).

Step 3: Calculate the linear mass density for the second string

Given the mass of the second string as \(16.0 \mathrm{~g}\) and the length as \(1.00 \mathrm{~m}\), the linear mass density can be calculated as follows: $$\mu_2 = \frac{16.0 \mathrm{~g}}{1.00 \mathrm{~m}} = 16.0 \mathrm{\frac gm}$$

Step 4: Calculate the first harmonic frequency for the second string

Using the formula for the fundamental frequency and the given tension of \(81.0 \mathrm{~N}\), the vibrating length of the second string as \(0.65 \mathrm{~m}\), and the calculated linear mass density: $$f_2 = \frac{1}{2(0.65 \mathrm{~m})} \sqrt{\frac{81.0 \mathrm{~N}}{16.0 \mathrm{\frac gm}}}$$ Converting grams to kilograms (to keep SI units): $$f_2 = \frac{1}{2(0.65 \mathrm{~m})} \sqrt{\frac{81.0 \mathrm{~N}}{0.016 \mathrm{\frac kgm}}} = 48.60 \mathrm{Hz}$$ So, the frequency of the first harmonic for the second string is approximately \(48.60 \mathrm{Hz}\). To answer both parts of the question: a) The frequency of the first harmonic for the first string is approximately \(61.61 \mathrm{Hz}\). b) The frequency of the first harmonic for the second string is approximately \(48.60 \mathrm{Hz}\).

Key Concepts

Linear Mass Density

Linear mass density, often symbolized as \( \mu \), is a property of strings that measures the mass per unit length. It is a critical parameter in determining the behavior of a string when it vibrates and can be calculated by dividing the string's total mass by its length. In our exercise example, for the first string, given a mass of 10.0 grams and length of 1.00 meter, the linear mass density was found to be \( 10.0 \, \text{g/m} \).

Understanding the concept of linear mass density is essential because it directly affects the frequency of the vibrations: a string with higher mass density will vibrate at a lower frequency, while a lighter string will vibrate at a higher frequency. This principle was observed in our exercise when comparing the two strings, where the heavier 16.0 grams string had a lower fundamental frequency.

Fundamental Frequency

The fundamental frequency, also known as the first harmonic, is the lowest frequency at which a string can vibrate. This frequency is crucial in music and engineering as it defines the pitch of the note that the string produces when it's plucked, struck, or bowed. The formula to calculate the fundamental frequency takes into account the tension in the string, its length, and its linear mass density:

\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]
where \( f \) is the fundamental frequency, \( L \) is the vibrating length of the string, \( T \) is the tension in the string, and \( \mu \) is the linear mass density. Applying this formula to the exercise provided, we calculated the fundamental frequencies for two guitar strings under the same tension but with different linear mass densities.

Vibrating String

A vibrating string is a foundational example of wave phenomena in physics. When a string is plucked, it vibrates in a pattern that can be broken down into a series of harmonics or overtones. The most basic of these is the fundamental mode, where the string vibrates in a single segment and produces the lowest frequency. However, the string can also vibrate in more complex patterns or higher harmonics, dividing into multiple segments and producing higher pitched tones.

For strings fixed at both ends, like those on a guitar, the vibration patterns are standing waves with nodes at the ends, where the displacement is always zero, and antinodes at points of maximum amplitude. This pattern is influenced by the nature of the string, including its tension and linear mass density, altering the speed of the wave and thus the pitch of the note.

Tension in Strings

Tension in strings is the force that stretches the string and is a key factor in defining how it vibrates. Guitar players are intimately familiar with the impact of tension, as tuning a guitar involves adjusting the tension until the desired pitch is produced. The relationship between tension in a string and the frequency of the vibration is directly proportional—increasing the tension leads to higher frequencies, and vice versa.

The equation for the fundamental frequency shows this dependence explicitly, highlighting that for a given linear mass density and string length, the tension is what primarily determines the pitch. In the exercise, the tension was constant at 81.0 N, but the frequencies of the two strings differed due to their differing linear mass densities.