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Chapter 15: Problem 73

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?

Short Answer

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

Step by step solution

01

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 \( T \).
02

Calculate Linear Mass Density

Convert the mass of the string from grams to kilograms: \( 14.4 \text{ g} = 0.0144 \text{ kg} \). Then calculate \( \mu = \frac{0.0144 \text{ kg}}{0.600 \text{ m}} = 0.024 \text{ kg/m} \).
03

Rearrange Formula for Tension

From the formula \( f = \frac{1}{2L} \sqrt{\frac{T}{\mu}} \), solve for \( T \): \( T = \mu (2Lf)^2 \).
04

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: \[ T = 0.024 \times (2 \times 0.600 \times 65.4)^2 \approx 102.7 \text{ N} \].
05

Calculate Tension for 73.4 Hz

Use the formula \( T = \mu (2Lf)^2 \) with \( f = 73.4 \text{ Hz} \): \[ T = 0.024 \times (2 \times 0.600 \times 73.4)^2 \approx 127.6 \text{ N} \].
06

Calculate Percent Increase in Tension

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

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Key Concepts

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

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.

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Most popular questions from this chapter

The upper end of a 3.80-m-long steel wire is fastened to the ceiling, and a 54.0-kg object is suspended from the lower end of the wire. You observe that it takes a transverse pulse 0.0492 s to travel from the bottom to the top of the wire. What is the mass of the wire?

A 0.800-m-long string with linear mass density \(\mu = 7.50\) g/m is stretched between two supports. The string has tension \(F\) and a standing-wave pattern (not the fundamental) of frequency 624 Hz. With the same tension, the next higher standing-wave frequency is 780 Hz. (a) What are the frequency and wavelength of the fundamental standing wave for this string? (b) What is the value of \(F\)?

A horizontal wire is stretched with a tension of 94.0 N, and the speed of transverse waves for the wire is 406 m/s. What must the amplitude of a traveling wave of frequency 69.0 Hz be for the average power carried by the wave to be 0.365 W?

A thin string 2.50 m in length is stretched with a tension of 90.0 N between two supports. When the string vibrates in its first overtone, a point at an antinode of the standing wave on the string has an amplitude of 3.50 cm and a maximum transverse speed of 28.0 m/s. (a) What is the string's mass? (b) What is the magnitude of the maximum transverse acceleration of this point on the string?

(a) A horizontal string tied at both ends is vibrating in its fundamental mode. The traveling waves have speed \(v\), frequency \(f\), amplitude \(A\), and wavelength \(\lambda\). Calculate the maximum transverse velocity and maximum transverse acceleration of points located at (i) \(x = \lambda/2\), (ii) \(x = \lambda /4\), and (iii) \(x = \lambda /8\), from the left-hand end of the string. (b) At each of the points in part (a), what is the amplitude of the motion? (c) At each of the points in part (a), how much time does it take the string to go from its largest upward displacement to its largest downward displacement?
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