Problem 87 - University Physics with Modern Physics

Question

Out of Tune. The B-string of a guitar is made of steel (density \(7800 \mathrm{kg} / \mathrm{m}^{3} ),\) is 63.5 \(\mathrm{cm}\) long, and has diameter 0.406 \(\mathrm{mm} .\) The fundamental frequency is \(f=247.0 \mathrm{Hz}\) . (a) Find the string tension. (b) If the tension \(F\) is changed by a small amount \(\Delta F,\) the frequency \(f\) changes by a small amount \(\Delta f .\) Show that $$\frac{\Delta f}{f}=\frac{\Delta F}{2 F}$$ (c) The string is tuned to a fundamental frequency of 247.0 Hz when its temperature is \(18.5^{\circ} \mathrm{C}\) . Strenuous playing can make the temperature of the string rise, changing its vibration frequency. Find \(\Delta f\) if the temperature of the string rises to \(29.5^{\circ} \mathrm{C}\) . The steel string has a Young's modulus of \(2.00 \times 10^{11} \mathrm{Pa}\) and a coefficient of linear expansion of \(1.20 \times 10^{-5}\left(\mathrm{C}^{\circ}\right)^{-1} .\) Assume that the temperature of the body of the guitar remains constant. Will the vibration frequency rise or fall?

Step-by-Step Solution

Answer

The tension in the string is 106.16 N. Frequency change due to temperature rise is 0.39 Hz, resulting in a frequency decrease.

Step 1: Calculate the Mass per Unit Length

First, find the cross-sectional area of the string using the diameter. The diameter is 0.406 mm which is 0.000406 m. The radius, \( r \), is half the diameter. Thus, \( r = 0.000203 \text{ m} \). The cross-sectional area \( A \) is given by \( A = \pi r^2 \). After calculating, \( A = \pi \times (0.000203)^2 = 1.29 \times 10^{-7} \text{ m}^2 \). The volume of the string can be calculated using the length and the cross-sectional area: \( V = A \times L \). Here, \( L = 0.635 \text{ m} \) (converted from 63.5 cm), so \( V = 1.29 \times 10^{-7} \times 0.635 = 8.20 \times 10^{-8} \text{ m}^3 \). The mass, \( m \), is the density times the volume: \( m = 7800 \times 8.20 \times 10^{-8} = 0.00640 \text{ kg} \). Therefore, the linear density \( \mu \) is \( \mu = \frac{m}{L} = \frac{0.00640}{0.635} = 0.010 \text{ kg/m} \).

Step 2: Calculate the String Tension

The fundamental frequency \( f \) of a string is related to its tension \( F \), length \( L \), and linear mass density \( \mu \) by the formula: \( f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \). Rearrange this formula to solve for the tension \( F \): \( F = 4L^2f^2\mu \). Substitute the known values: \( F = 4 \times (0.635)^2 \times (247)^2 \times 0.010 \). After calculations, \( F = 106.16 \text{ N} \).

Step 3: Derive the Frequency Change Formula

To show that \( \frac{\Delta f}{f} = \frac{\Delta F}{2F} \), start by differentiating the frequency equation: \( f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \) with respect to \( F \). This differentiation gives \( \frac{df}{dF} = \frac{1}{4L} \sqrt{\frac{1}{\mu F}} \). Thus, \( \Delta f \approx \frac{1}{4L} \sqrt{\frac{1}{\mu F}} \Delta F \). By dividing this by \( f \) and substituting back, \( \frac{\Delta f}{f} = \frac{\Delta F}{2F} \).

Step 4: Calculate Frequency Change Due to Temperature

The frequency change due to temperature change \( \Delta T \) is related to the change in tension \( \Delta F \). Due to thermal expansion, the change in length is \( \Delta L = \alpha L_0 \Delta T \) where \( \alpha = 1.20 \times 10^{-5} \underline{\phantom{xxx}}^{\circ}\text{C}^{-1} \). The change in tension can be estimated using the Young's modulus: \( \Delta F = \frac{E \Delta L}{L_0} \times A \), so \( \Delta F = E\alpha A \Delta T \). Substituting given values: \( \Delta F = 2.00 \times 10^{11} \times 1.20 \times 10^{-5} \times 1.29 \times 10^{-7} \times 11 = 0.34 \text{ N} \). With \( \frac{\Delta f}{f} = \frac{\Delta F}{2F} \), \( \Delta f = \frac{0.34}{2 \times 106.16} \times 247 = 0.39 \text{ Hz} \).

Step 5: Determine the Direction of Frequency Change

A rise in temperature causes the string to expand, decreasing the tension if the stretching is not compensated. As a result, the frequency will decrease.

Key Concepts

String Tension

String tension is a critical factor in wave mechanics, particularly for musical instruments like guitars. It refers to the force that stretches the strings and determines their vibration characteristics. In simple terms, it is the pull trying to make the string longer. The tension can be calculated using the formula: \[ F = 4L^2f^2\mu \] where:

• \( F \) is the tension in Newtons,
• \( L \) is the length of the string in meters,
• \( f \) is the frequency of the fundamental tone in Hertz,
• \( \mu \) is the linear mass density of the string in kilograms per meter.

Tension affects the pitch of the string: increasing the tension raises the pitch, while decreasing it lowers the pitch. A change in tension can result from several factors like temperature fluctuations or changes in string length.

Vibration Frequency

Vibration frequency is a term that refers to the number of oscillations a vibrating string makes in one second. It is measured in Hertz (Hz). In a fixed-length string like a guitar's, its vibration depends on:

• String tension, \( F \)
• String length, \( L \)
• Linear mass density, \( \mu \)

The fundamental frequency, the lowest frequency at which the string vibrates, is given by:\[ f = \frac{1}{2L} \sqrt{\frac{F}{\mu}} \]This equation tells us that:

• Higher tension or lighter strings increase frequency, resulting in higher pitches.
• Longer strings lead to lower frequencies, lowering pitches.

Understanding the interplay between tension, mass, and length is crucial to tuning musical instruments and predicting how temperature changes might affect these characteristics.

Thermal Expansion

Thermal expansion is the phenomenon where materials change in shape, area, and volume in response to temperature changes. For a guitar string, thermal expansion means that as the string gets warmer, it lengthens. The change in length due to a temperature change \( \Delta T \) can be determined using the coefficient of linear expansion \( \alpha \):\[ \Delta L = \alpha L_0 \Delta T \]where:

• \( \Delta L \) is the change in length,
• \( \alpha \) is the coefficient of linear expansion,
• \( L_0 \) is the original length of the string,
• \( \Delta T \) is the change in temperature.

When a string experiences thermal expansion:

• It generally becomes longer.
• If not balanced by increased tension, frequency decreases.

This needs to be understood and managed to keep instruments properly tuned under varying thermal conditions.

Young's Modulus

Young's modulus is a measure of the elasticity of a material, expressed as a ratio of stress to strain. It tells us how easily a material can be stretched or compressed. For the steel guitar string described, Young's modulus \( E \) is crucial for determining how changes in temperature affect tension:

• Young's modulus is high for steel, indicating it is relatively stiff and does not deform easily.
• Due to this rigidity, a small change in length (from thermal expansion) results in considerable stress and strain.

Young's modulus is factored into the equation:\[ \Delta F = \frac{E \Delta L}{L_0} \times A \]where:

• \( \Delta F \) is the change in tension,
• \( A \) is the cross-sectional area of the string.

Instruments with components made from materials with high Young's modulus tend to maintain their tuning better despite environmental changes.