Problem 82 - University Physics with Modern Physics

Question

Tuning an Instrument. 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 \(\mathrm{m}\) long and has a mass of 14.4 \(\mathrm{g} .\) (a) With what tension must the musician stretch it? (b) What percent increase in tension is needed to increase the frequency from 65.4 \(\mathrm{Hz}\) to 73.4 \(\mathrm{Hz}\) , corresponding to a rise in pitch from \(\mathrm{C}\) to \(\mathrm{D} ?\)

Step-by-Step Solution

Answer

The tension needed is 100.65 N, and a 26.7% increase in tension is needed.

Step 1: Calculate the Linear Density of the String

To find the linear density (\( \mu \)) of the string, we need to divide the mass by the length. Convert the mass from grams to kilograms: \( 14.4\ \text{g} = 0.0144\ \text{kg} \). The linear density is computed as follows: \( \mu = \frac{0.0144}{0.600} = 0.024\ \text{kg/m} \).

Step 2: Use Frequency Formula to Find Tension

The formula relating tension (\(T\)), frequency (\(f\)), length (\(L\)), and linear density (\(\mu\)) is: \(f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}\). Rearrange the formula to solve for the tension: \(T = (2L f)^2 \mu\). Substitute \(L = 0.600\ \text{m}, f = 65.4\ \text{Hz}, \text{and } \mu = 0.024\ \text{kg/m}\): \(T = (2 \times 0.600 \times 65.4)^2 \times 0.024 = 100.65\ \text{N}\).

Step 3: Calculate New Tension for Increased Frequency

Substitute the new frequency \(f = 73.4\ \text{Hz}\) into the tension equation: \(T' = (2 \times 0.600 \times 73.4)^2 \times 0.024\). This calculates to \(T' = 127.54\ \text{N}\).

Step 4: Calculate Percent Increase in Tension

Find the percent increase in tension using the formula: \(\text{Percent Increase} = \frac{T' - T}{T} \times 100\). Substitute \(T = 100.65\ \text{N} \) and \(T' = 127.54\ \text{N}\) into the formula: \(\text{Percent Increase} = \frac{127.54 - 100.65}{100.65} \times 100 \approx 26.7\%\).

Key Concepts

String Tension

String tension is a critical element in determining the sound produced by a stringed instrument. In simple terms, tension refers to how tightly a string is stretched. Here's why it's important:

• Tension affects the vibration frequency of the string - the tighter the string, the higher the frequency.
• More tension leads to a higher pitch, which is essential for tuning the instrument.

In the context of wave mechanics, tension is related to how fast waves travel along the string. When you increase the tension, the speed of the wave increases, leading to a higher sound frequency. For a musician aiming to hit the perfect pitch on an instrument, adjusting the string tension is vital. To calculate the necessary tension for a string when tuning, we use the formula: \[ T = (2L f)^2 \mu \] where \( L \) is the length of the vibrating portion of the string, \( f \) is the fundamental frequency, and \( \mu \) is the linear density.

Fundamental Frequency

The fundamental frequency is the lowest frequency at which a string can vibrate. It's also known as the first harmonic. This frequency determines the basic pitch of the note produced by the string.
The fundamental frequency is crucial because:

• It serves as the base note or the primary tone.
• Musicians adjust their instruments to match this frequency for proper tuning.

For example, when a musician tunes the C-string to 65.4 Hz, that frequency becomes the fundamental frequency. It is important to note that while the fundamental frequency is observed, the string also vibrates at various other frequencies called overtones, but these do not affect the main pitch. The fundamental frequency is calculated using the formula:\[ f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \]where \( L \) is the vibrating length, \( T \) is the tension, and \( \mu \) is the linear density of the string.

Linear Density

Linear density is a measure of how much mass is associated with a unit length of the string, denoted as \( \mu \). It's calculated by taking the total mass of the string and dividing it by its length.
This concept is significant for a few reasons:

• The linear density can affect both the tension and the fundamental frequency.
• A higher linear density means more mass distributed across the string, which usually results in a lower pitch.

In practical terms, linear density is crucial for determining how a string responds to tension and what kind of frequency it will produce. For the string in the problem, we calculated linear density as:\[ \mu = \frac{0.0144}{0.600} = 0.024 \text{ kg/m} \]A correct calculation of linear density helps to ensure that instruments can be accurately tuned to the desired pitch.

Frequency and Pitch

Frequency and pitch are closely related, as frequency is a numerical expression of pitch. In musical terms, when the frequency of a string's vibration increases, the pitch of the note produced becomes higher.
The relationship between frequency and pitch is important for several reasons:

• Tuning instruments involves adjusting the frequency to match a desired pitch.
• Musicians need to understand this relationship to achieve harmonious notes.

In the exercise, moving from a frequency of 65.4 Hz to 73.4 Hz represents a change in pitch from a C to a D note. This shift requires the string to be retensioned appropriately. It's helpful to remember that while frequency measures the actual number of cycles per second, pitch is more of a perceptual quality that tells us how high or low a note sounds. Both concepts are vital in ensuring instruments produce the correct sounds during a performance.