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Chapter 16: Problem 85

Suppose the linear density of the A string on a violin is \(7.8 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). A wave on the string has a frequency of \(440 \mathrm{~Hz}\) and a wavelength of \(65 \mathrm{~cm}\). What is the tension in the string?

Short Answer

Expert verified
The tension in the string is approximately 63.12 N.

Step by step solution

01

Identify Given Values and Units

First, let's identify and convert the given values to standard units. We have a linear density \( \mu = 7.8 \times 10^{-4} \ \text{kg/m} \), frequency \( f = 440 \ \text{Hz} \), and wavelength \( \lambda = 65 \ \text{cm} = 0.65 \ \text{m} \).
02

Calculate Wave Speed

The speed \( v \) of the wave on the string can be found using the formula \( v = f \lambda \). Substitute the known values: \[ v = 440 \ \text{Hz} \times 0.65 \ \text{m} = 286 \ \text{m/s} \].
03

Relate Wave Speed to Tension

The wave speed on a string is also related to the tension \( T \) and the linear density \( \mu \) by the formula \( v = \sqrt{\frac{T}{\mu}} \). We rearrange this formula to find tension: \( T = \mu v^2 \).
04

Calculate Tension

Now substitute \( \mu = 7.8 \times 10^{-4} \ \text{kg/m} \) and \( v = 286 \ \text{m/s} \) into the tension formula: \[ T = 7.8 \times 10^{-4} \ \text{kg/m} \times (286 \ \text{m/s})^2 \].
05

Compute Final Value for Tension

Calculate the tension: \[ T = 7.8 \times 10^{-4} \times 81956 = 63.12 \ \text{N} \].

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Key Concepts

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

Linear Density
Linear density is a key factor in wave mechanics, especially when dealing with waves on strings. It refers to the mass per unit length of the string. In our example, the linear density, denoted as \( \mu \), of the violin string is \(7.8 \times 10^{-4} \ \text{kg/m}\). This means that for every meter of the string, the mass is \(0.00078\) kilograms.

Understanding linear density is crucial because it directly affects how waves travel along the string. A higher linear density typically means a slower wave speed given the same tension, as the string is heavier and takes more force to move.
  • Units: Typically measured in kilograms per meter (kg/m).
  • Factors: Influenced by the material of the string and its thickness.
  • Role: Affects the wave speed and can determine the note that a string plays when plucked.
Frequency
Frequency is the number of oscillations a wave makes per second, measured in hertz (Hz). In music, frequency determines the pitch of a note. More oscillations mean a higher pitch. For the violin string discussed, the frequency is \(440\ \text{Hz}\), which corresponds to the musical note 'A'.

This value means the string vibrates 440 times each second.
  • Relationship: Frequency is related to wave speed through the formula \( f = \frac{v}{\lambda} \), where \(v\) is wave speed and \(\lambda\) is wavelength.
  • Pitch: High frequency = high pitch; low frequency = low pitch.
  • Applications: In tuning musical instruments and understanding sound physics.
Frequency helps to set the fundamental tone of the string鈥檚 vibration, impacting how we perceive the sound.
Wave Speed
Wave speed on a string is determined by both the tension in the string and its linear density. For a wave on our violin string, we calculated the wave speed \( v \) to be \(286 \ \text{m/s}\). This calculation used the formula \( v = f \lambda \), combining both frequency and wavelength.

Wave speed informs us how quickly a wave travels down the length of the string.
  • Equation: \( v = f \lambda \), where \( f \) is frequency and \( \lambda \) is wavelength.
  • Impact: Faster wave speeds lead to higher pitched notes, assuming other factors are controlled.
  • Relation: Wave speed also connects to string tension through \( v = \sqrt{\frac{T}{\mu}} \).
This speed is essential for correctly producing musical notes as it influences how energy is transmitted through the string.
String Tension
String tension is the force exerted along a string, which significantly affects wave speed and vibration characteristics. In our violin example, string tension was found using the formula \( T = \mu v^2 \), resulting in a tension \( T \) of \(63.12 \ \text{N}\).

In practical terms, tension influences both pitch and wave speed. Higher tension results in faster waves and, thus, higher pitch sounds.
  • Formula: Rearranged from wave speed formula: \( T = \mu v^2 \).
  • Influences: Affects wave speed and vibration frequency on stringed instruments.
  • Sound Production: Critical in tuning strings to achieve desired musical pitch.
By adjusting the tension, musicians can refine the notes produced, making it essential for accurate musical performance and instrument setup.

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Most popular questions from this chapter

As the drawing shows, one microphone is located at the origin, and a second microphone is located on the \(+y\) axis. The microphones are separated by a distance of \(D=1.50 \mathrm{~m}\). A source of sound is located on the \(+x\) axis, its distances from microphones 1 and 2 being \(L_{1}\) and \(L_{2}\), respectively. The speed of sound is \(343 \mathrm{~m} / \mathrm{s}\). The sound reaches microphone 1 first, and then, 1.46 ms later, it reaches microphone 2 . Find the distances \(L_{1}\) and \(L_{2}\)

In a discussion person \(\mathrm{A}\) is talking \(1.5 \mathrm{~dB}\) louder than person \(\mathrm{B},\) and person \(\mathrm{C}\) is talking \(2.7 \mathrm{~dB}\) louder than person \(\mathrm{A}\). What is the ratio of the sound intensity of person \(\mathrm{C}\) to the sound intensity of person \(\mathrm{B} ?\)

A hunter is standing on flat ground between two vertical cliffs that are directly opposite one another. He is closer to one cliff than to the other. He fires a gun and, after a while, hears three echoes. The second echo arrives \(1.6 \mathrm{~s}\) after the first, and the third echo arrives \(1.1 \mathrm{~s}\) after the second. Assuming that the speed of sound is \(343 \mathrm{~m} / \mathrm{s}\) and that there are no reflections of sound from the ground, find the distance between the cliffs.

A longitudinal wave with a frequency of \(3.0 \mathrm{~Hz}\) takes \(1.7 \mathrm{~s}\) to travel the length of a \(2.5-\mathrm{m}\) Slinky (see Figure \(16-3\) ). Determine the wavelength of the wave.

As a prank, someone drops a water-filled balloon out of a window. The balloon is released from rest at a height of \(10.0 \mathrm{~m}\) above the ears of a man who is the target. Then, because of a guilty conscience, the prankster shouts a warning after the balloon is released. The warning will do no good, however, if shouted after the balloon reaches a certain point, even if the man could react infinitely quickly. Assuming that the air temperature is \(20^{\circ} \mathrm{C}\) and ignoring the effect of air resistance on the balloon, determine how far above the man's ears this point is.
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