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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 \(\mathrm{D} ?\)

Short Answer

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

Step by step solution

01

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.
02

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 \( \mu = \frac{0.0144}{0.600} = 0.024 \, \text{kg/m} \).
03

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.
04

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.
05

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 \( \frac{(148.10976 - 120.02496)}{120.02496} \times 100\% = 23.42\% \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A trumpet player is tuning his instrument by playing an A note simultaneously with the first-chair trumpeter, who has perfect pitch. The first-chair player's note is exactly 440 Hz, and 2.8 beats per second are heard. What are the two possible frequencies of the other player's note?

\(\bullet\) If an earthquake wave having a wavelength of 13 \(\mathrm{km}\) causes the ground to vibrate 10.0 times each minute, what is the speed of the wave?

You're standing between two speakers that are driven by the same amplifier and are emitting sound waves with frequency 229 Hz. The two speakers are facing each other, 15 meters apart. (a) You begin walking away from one speaker toward the other one, and as you walk, you hear what sounds like beats, with a frequency of 2.50 Hz. How fast are you walking? (b) If the frequency of the sound emitted by the speakers increases to 573 Hz and you continue to walk at the same speed, what frequency of beats will you hear? [Hint: You can model this situation as a tube open at both ends; alternatively, you can treat it as a Doppler effect problem.]

\(\bullet\) The fundamental frequency of a pipe that is open at both ends is 594 Hz. (a) How long is this pipe? If one end is now closed, find (b) the wavelength and (c) the frequency of the new fundamental.

Moving source vs. moving listener. (a) A sound source producing 1.00 kHz waves moves toward a stationary listener at one-half the speed of sound. What frequency will the listener hear? (b) Suppose instead that the source is stationary and the listener moves toward the source at one-half the speed of sound. What frequency does the listener hear? How does your answer compare with that in part (a)? Did you expect to get the same answer in both cases? Explain on physical grounds why the two answers differ.

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