Problem 18 - Fundamentals of Physics

Question

The heaviest and lightest strings on a certain violin have linear densities of \(3.0\) and \(0.29 \mathrm{~g} / \mathrm{m}\). What is the ratio of the diameter of the heaviest string to that of the lightest string, assuming that the strings are of the same material?

Step-by-Step Solution

Answer

The ratio of the diameters is approximately 3.217.

Step 1: Understanding Materials' Density

Since the strings are made from the same material, they have the same mass density (\(\rho\)). This density is defined as \(\rho = \frac{m}{V}\) where \(m\) is mass and \(V\) is volume. For strings, linear density (\( \mu \)) is used, which is mass per unit length: \(\mu = \frac{m}{L}\). Since \(\mu\) directly depends on the string's diameter \(d\), we relate it to \(\rho\) and \(d\).

Step 2: Express Linear Density in Terms of Diameter

For a string with circular cross-section, the volume \(V\) for length \(L\) is \(V = A \cdot L \), where \( A = \pi \left( \frac{d}{2} \right)^2 \). The linear density is \(\mu = \frac{m}{L} = \frac{ \rho \cdot V}{L} = \rho \cdot \pi \left( \frac{d}{2} \right)^2 = \rho \cdot \frac{\pi d^2}{4}\).

Step 3: Set Up the Ratio of Diameters

Given two strings with linear densities \(\mu_1 = 3.0 \, \text{g/m}\) and \(\mu_2 = 0.29 \, \text{g/m}\), their diameters \(d_1\) and \(d_2\) can be related. Using the formula \(\mu = \rho \cdot \frac{\pi d^2}{4}\), we can write \(\mu_1 = \rho \cdot \frac{\pi d_1^2}{4}\) and \(\mu_2 = \rho \cdot \frac{\pi d_2^2}{4}\).

Step 4: Calculate Diameter Ratio

To find the ratio of the diameters, divide the two linear density equations: \(\frac{\mu_1}{\mu_2} = \frac{d_1^2}{d_2^2}\). Solving for the diameter ratio, we have \(\frac{d_1}{d_2} = \sqrt{\frac{\mu_1}{\mu_2}} = \sqrt{\frac{3.0}{0.29}}\).

Step 5: Execute Calculation

Perform the calculation: \(\frac{d_1}{d_2} = \sqrt{\frac{3.0}{0.29}} \approx \sqrt{10.3448} \approx 3.217\). This is the ratio of the diameters of the heaviest string to the lightest string.

Key Concepts

Linear Density

Linear density, represented as \( \mu \), is a measure of how much mass a string has per unit length. It is essentially the amount of mass that spans over one meter of the string. The higher the linear density, the heavier the string. For strings like those on a violin, linear density plays a crucial role in the sound they produce. It can be calculated using the formula:\[ \mu = \frac{m}{L} \]where \( m \) is the mass and \( L \) is the length of the string. This directly ties into the string's diameter because a thicker string holds more mass.

• Linear Density increases with more mass in a given length.
• It affects the flexibility and tension requirement of the string.
• Different linear densities lead to diverse sound characteristics.

Material Density

Material density (\( \rho \)) is a fundamental property of the material, defining how much mass exists within a given volume. Since the strings in the exercise are made from the same material, they share the same density. This uniformity simplifies calculations when comparing strings of identical materials.
Material density is expressed as:\[ \rho = \frac{m}{V} \]Where \( V \) is the volume. In practical terms, it tells us how compact the material structure is. Lower density might imply a material is less massive per unit volume, potentially being more flexible.

• Material Density is constant for materials of the same kind.
• It helps determine how mass is distributed over the string's volume.
• Vital in calculating derived properties like linear density.

Diameter Ratio

The diameter ratio of two objects offers insight into their relative thicknesses. In this exercise, understanding the diameter ratio of the violin strings provides information on their mechanical and acoustic properties. The diameter influences both the string's linear density and its vibrational characteristics.
To find this ratio, use the relationship:\[ \frac{d_1}{d_2} = \sqrt{\frac{\mu_1}{\mu_2}} \]Here, \( \mu_1 \) and \( \mu_2 \) represent the linear densities of the strings. The larger the ratio, the thicker the heavier string compared to the lighter one, affecting its ability to produce lower pitches.

• Larger diameter implies a string can be less flexible and carry more tension.
• Influences how sound waves propagate along the string.
• Informs about the design and tuning of musical instruments.

String Mass

The string mass is the total mass of a violin string, an important factor dictated by both linear density and overall length. To find the mass, the formula for linear density comes into play:\[ m = \mu \times L \]Where \( m \) is mass, \( \mu \) is the linear density, and \( L \) is the length. Hence, even a string with small diameter but high linear density can result in a significant mass, influencing its tension and vibrational frequency.
Understanding string mass helps in several areas:

• Determines the amount of tension needed to produce a particular tone.
• Indicates how durable and strong the string needs to be under tension.
• Essential for adjusting brevity and sonority of string notes.