Problem 66 - College Physics

Question

Tuning a cello. A cellist tunes the C-string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.600 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g} .\) (a) With what tension must she stretch that portion of the string? (b) What percentage increase in tension is needed to increase the frequency from 65.4 Hz to 73.4 Hz, corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

Step-by-Step Solution

Answer

(a) Tension is 120.02 N; (b) Percentage increase in tension is 23.42%.

Step 1: Understand the Formula for Frequency of a String

The fundamental frequency of a vibrating string is given by the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), where \( f \) is the frequency, \( L \) is the length of the string, \( T \) is the tension in the string, and \( \mu \) is the linear mass density of the string.

Step 2: Calculate Linear Mass Density \( \mu \)

The linear mass density \( \mu \) is the mass per unit length, calculated as \( \mu = \frac{m}{L} \). Given \( m = 14.4 \) g = 0.0144 kg and \( L = 0.600 \) m, we find \( \mu = \frac{0.0144}{0.600} = 0.024 \, \text{kg/m} \).

Step 3: Calculate Tension for Frequency 65.4 Hz

Using the formula for frequency, \( T = (2L \cdot f)^2 \cdot \mu \). Substituting \( f = 65.4 \) Hz, \( L = 0.600 \) m, and \( \mu = 0.024 \) kg/m, we get \( T = (2 \cdot 0.600 \cdot 65.4)^2 \cdot 0.024 = 120.02496 \) N.

Step 4: Calculate Tension for Frequency 73.4 Hz

Substituting \( f = 73.4 \) Hz in the formula, \( T = (2L \cdot f)^2 \cdot \mu \). We get \( T = (2 \cdot 0.600 \cdot 73.4)^2 \cdot 0.024 = 148.10976 \) N.

Step 5: Determine the Percentage Increase in Tension

The percentage increase in tension is given by \( \frac{(T_{new} - T_{old})}{T_{old}} \times 100\% \). Substituting \( T_{new} = 148.10976 \) N and \( T_{old} = 120.02496 \) N, we find \( \frac{(148.10976 - 120.02496)}{120.02496} \times 100\% = 23.42\% \).

Key Concepts

String tension calculation

Understanding the tension in a string is crucial for controlling the pitch of musical instruments like a cello. The tension refers to the pulling force applied along the string, which affects its vibration and resulting sound.In the problem provided, we're tasked with finding the tension needed for specific string frequencies. We use the formula for tension derived from the wave frequency formula: - Start with the known frequency formula: \[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]- Rearrange to solve for the string tension \( T \): \[ T = (2Lf)^2 \mu \]Using this equation, you can determine the tension needed to achieve the desired frequency, taking into account the string’s length \( L \) and its linear mass density \( \mu \). Ensuring the correct tension keeps the string in tune and maintains sound quality.

Frequency of a string

The frequency of a string is essentially how fast it vibrates when plucked or played. It's directly connected to the perceived pitch: higher frequencies sound higher in pitch.When solving problems involving string frequency, remember:- The fundamental frequency, represented as \( f \), is the lowest natural frequency at which a string vibrates.- It is determined by the length of the string \( L \), its tension \( T \), and its linear mass density \( \mu \). To adjust the pitch from one note to another, as in the case from C to D in our example, you might need to change the tension, altering the frequency of the string's vibration. This can be especially important in musical contexts, where precise tuning is crucial for harmonious play.

Linear mass density

Linear mass density is a key factor in understanding how a string behaves when vibrating. It is defined as the mass per unit length of a string, denoted by \( \mu \).To calculate linear mass density:- Use the formula: \( \mu = \frac{m}{L} \)- Where \( m \) is the mass of the string, and \( L \) is its vibrating length.In our example, the cello string has a mass of 14.4 g and a length of 0.6 m, leading to \( \mu = 0.024 \, \text{kg/m} \).The value of \( \mu \) affects both the frequency and the tension needed for specific pitches. Strings with higher linear mass densities will generally require more tension to achieve the same frequency as strings with lower linear mass densities.

Wave frequency formula

The wave frequency formula is fundamental to calculations involving vibrating strings in physics. This formula ties together the string's physical properties to determine its vibrational behavior.The formula used in our example:- \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \)- Each component is crucial: - \( f \) stands for frequency, the rate of vibration. - \( L \) is the length of the vibrating string. - \( T \) represents the tension in the string. - \( \mu \) is the linear mass density.This formula shows that increasing tension or reducing linear mass density raises the frequency, thereby increasing the pitch. It's an essential equation in both physics and music for controlling and predicting the behavior of string instruments.