Problem 73 - University Physics with Modern Physics

Question

A musician tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g. (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from C to D?

Step-by-Step Solution

Answer

(a) The tension is approximately 102.7 N. (b) The percent increase in tension is about 24.2%.

Step 1: Identify the Formula

The fundamental frequency of a vibrating string is given by the formula: \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \). Here, \( f \) is the frequency, \( L \) is the length of the string, \( T \) is the tension, and \( \mu \) is the linear mass density \( \left( \mu = \frac{m}{L} \right) \). We need to calculate \( T \).

Step 2: Calculate Linear Mass Density

Convert the mass of the string from grams to kilograms: \( 14.4 \text{ g} = 0.0144 \text{ kg} \). Then calculate \( \mu = \frac{0.0144 \text{ kg}}{0.600 \text{ m}} = 0.024 \text{ kg/m} \).

Step 3: Rearrange Formula for Tension

From the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), solve for \( T \): \( T = \mu (2Lf)^2 \).

Step 4: Calculate Initial Tension for 65.4 Hz

Substitute \( f = 65.4 \text{ Hz} \), \( L = 0.600 \text{ m} \), and \( \mu = 0.024 \text{ kg/m} \) into the equation: \[ T = 0.024 \times (2 \times 0.600 \times 65.4)^2 \approx 102.7 \text{ N} \].

Step 5: Calculate Tension for 73.4 Hz

Use the formula \( T = \mu (2Lf)^2 \) with \( f = 73.4 \text{ Hz} \): \[ T = 0.024 \times (2 \times 0.600 \times 73.4)^2 \approx 127.6 \text{ N} \].

Step 6: Calculate Percent Increase in Tension

The percent increase in tension is calculated as: \( \frac{127.6 - 102.7}{102.7} \times 100\% \approx 24.2\% \).

Key Concepts

Fundamental Frequency

Understanding the fundamental frequency is crucial when analyzing musical instruments like strings. The fundamental frequency refers to the lowest frequency at which a string vibrates. It is the basic note produced when the string is plucked or bowed. The frequency is determined by three factors:

• Length of the string: Longer strings produce lower frequencies.
• Tension in the string: Greater tension increases the frequency.
• Linear mass density of the string: Strings with greater mass per unit length vibrate at lower frequencies.

Notably, the fundamental frequency equation for a string is given by:\[f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\]where \( f \) is the frequency, \( L \) is the string length, \( T \) is the tension, and \( \mu \) is the linear mass density. Understanding these relationships can help musicians tune their instruments to the desired pitch.

Wave Equation

The wave equation provides a mathematical description of how waves propagate through a medium. In the context of strings, it helps determine the frequencies at which a string vibrates. For a wave traveling along a string, the relationship between the tension \( T \), linear mass density \( \mu \), and the speed \( v \) of the wave is given by:

• The speed of the wave on the string is \( v = \sqrt{\frac{T}{\mu}} \).
• The fundamental frequency and wave speed are related as \( f = \frac{v}{2L} \).

Thus, the wave equation ties together several physical properties of the string, allowing precise calculation of vibration and hence musical tone. This relationship is foundational in musical acoustics and physics.

Linear Mass Density

Linear mass density, often denoted by \( \mu \), plays a significant role in understanding string vibration. It is defined as the mass per unit length of the string:\[\mu = \frac{m}{L}\]where \( m \) is the mass of the string and \( L \) is its length. Linear mass density directly affects both the tension needed for a certain frequency and the wave speed on the string.

• A higher \( \mu \) will result in a slower wave speed and lower frequencies if other factors remain constant.
• Typically, for musical instruments, strings need an optimal mass density to ensure they produce the required tonal quality.

Understanding linear mass density helps musicians in selecting the right strings for desired pitches, enhancing tuning precision. This concept is crucial when calculating the physical properties and expected sound characteristics of musical instruments.