Problem 74 - College Physics

Question

A tight uniform string with a length of \(1.80 \mathrm{~m}\) is tied down at both ends and placed under a tension of \(100 \mathrm{~N}\). When it vibrates in its third harmonic (draw a sketch), the sound given off has a frequency of \(75.0 \mathrm{~Hz}\). What is the mass of the string?

Step-by-Step Solution

Answer

The mass of the string is approximately 0.02214 kg.

Step 1: Understand the concept of harmonic frequencies

When a string vibrates in its harmonics, it forms standing waves. The frequency of the nth harmonic is given by the formula \( f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \), where \( n \) is the harmonic number, \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear mass density.

Step 2: Setup the equation for the third harmonic

For the third harmonic ( \( n = 3 \)), the frequency is \( f_3 = 75.0 \mathrm{~Hz} \). Substitute into the formula for harmonics \( f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \):\[75.0 = \frac{3}{2 \times 1.80} \sqrt{\frac{100}{\mu}}.\]

Step 3: Solve for the linear mass density \( \mu \)

Rearrange the equation to isolate \( \mu \):\[75.0 = \frac{3}{3.6} \sqrt{\frac{100}{\mu}}.\]Simplify the fraction:\[75.0 = \frac{5}{6} \sqrt{\frac{100}{\mu}}.\]Square both sides to eliminate the square root:\[(75.0)^2 = \left( \frac{5}{6} \right)^2 \times 100 \div \mu.\]Calculate:\[5625 = \frac{25}{36} \times \frac{100}{\mu}.\]Isolate \( \mu \):\[\mu = \frac{25 \times 100}{36 \times 5625}.\]

Step 4: Calculate the value of \( \mu \)

Compute the linear mass density \( \mu \):\[\mu = \frac{2500}{202500} \approx 0.012345679.\]So, \( \mu \approx 0.0123 \mathrm{~kg/m} \).

Step 5: Calculate the mass of the string

Use the relationship \( \mu = \frac{m}{L} \) to find the mass \( m \):\[0.0123 = \frac{m}{1.80}.\]Solve for \( m \):\[m = 0.0123 \times 1.80.\]Calculate:\[m = 0.02214 \mathrm{~kg}.\]

Step 6: Conclusion

The mass of the string is approximately \(0.02214 \mathrm{~kg}\), or \(22.14\,\text{grams}\).

Key Concepts

Harmonic Frequencies

The concept of harmonic frequencies is crucial when understanding how strings vibrate. A harmonic frequency occurs when the natural frequency of vibration of the string matches an external frequency that causes the object to vibrate. In simpler terms, this means the string is vibrating in a way that creates whole-number multiples of a fundamental frequency. This results in distinct patterns of vibration known as harmonics.
In mathematical terms, the frequency of any harmonic on a string is given by the formula: \[ f_n = \frac{n}{2L} \sqrt{\frac{T}{\mu}} \] Here:

• \(f_n\) is the frequency of the nth harmonic.
• \(n\) represents which harmonic (e.g., first, second, third).
• \(L\) is the length of the string.
• \(T\) is the tension in the string.
• \(\mu\) is the linear mass density of the string.

Each harmonic corresponds to a different mode of vibration, giving the string its unique "voice." Understanding harmonics is fundamental to exploring sound physics and string instruments.

Standing Waves

Standing waves are patterns that emerge on strings when they vibrate, typically seen when the string is tied down at both ends like a guitar string. They are characterized by nodes and antinodes. A node is a point that remains at rest, while an antinode is a point that vibrates with the maximum amplitude.
For standing waves to form, the length of the string must accommodate whole or half wavelengths of the vibrating frequency. Each harmonic corresponds to a specific standing wave pattern on the string. As an example, in the third harmonic as calculated in the original exercise, the string is divided into three distinct parts with nodes at both ends and a node in the middle, bridged by antinodes. This harmonic behavior allows a string not only to vibrate at a fundamental frequency but also at its overtones, which are multiples of the fundamental frequency. Standing waves help musicians tune their instruments by providing the right kind of resonance, as each harmonic will emphasize different frequencies.

Linear Mass Density

Linear mass density, often symbolized as \(\mu\), is a measure of mass per unit length of a string. It is a critical factor in determining how a string will vibrate and at what frequencies. In essence, it's about how much mass of the string there is over a given length.
This concept is central to calculating harmonic frequencies because it affects the velocity at which waves can move through the string. The formula that relates the linear mass density with the string's mass \(m\) and length \(L\) is: \[ \mu = \frac{m}{L} \] Where:

• \(m\) is the total mass of the string.
• \(L\) is the total length of the string.

In practical terms, a string with lower linear mass density will have less inertia, allowing it to vibrate more easily at higher frequencies, while a string with higher mass density will vibrate at lower frequencies. Understanding the linear mass density is essential in physics and engineering fields where accurate modeling of wave propagation is required.