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The linear density of a string is \(1.6 \times 10^{-4} \mathrm{~kg} / \mathrm{m}\). A transverse wave on the string is described by the equation $$ y=(0.021 \mathrm{~m}) \sin \left[\left(2.0 \mathrm{~m}^{-1}\right) x+\left(30 \mathrm{~s}^{-1}\right) t\right] $$ What are (a) the wave speed and (b) the tension in the string?

Short Answer

Expert verified
The wave speed is 15 m/s, and the tension is 3.6 N.

Step by step solution

01

Identify wave equation components

The given wave equation is \( y = (0.021 \, \text{m}) \sin\left[(2.0 \, \text{m}^{-1}) x + (30 \, \text{s}^{-1}) t \right] \). From this equation, we determine the angular wave number \( k = 2.0 \, \text{m}^{-1} \) and the angular frequency \( \omega = 30 \, \text{s}^{-1} \).
02

Calculate wave speed

The wave speed \( v \) can be found using the relationship \( v = \frac{\omega}{k} \). Substituting the known values, \( v = \frac{30 \, \text{s}^{-1}}{2.0 \, \text{m}^{-1}} = 15 \, \text{m/s} \). Thus, the wave speed is 15 m/s.
03

Use wave speed to find tension

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

Substitute values to find tension

Given that the linear density \( \mu = 1.6 \times 10^{-4} \, \text{kg/m} \) and \( v = 15 \, \text{m/s} \), substitute these into the tension formula: \( T = 1.6 \times 10^{-4} \, \text{kg/m} \times (15 \, \text{m/s})^2 = 3.6 \, \text{N} \).
05

Conclusion: Results for wave speed and tension

The wave speed is 15 m/s and the tension in the string is 3.6 N.

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Key Concepts

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

Wave Speed
Wave speed is an essential concept in wave mechanics, as it tells us how fast a wave travels through a medium. It's like the car speedometer but for waves! When dealing with a wave on a string, the wave speed can be calculated using the relationship between angular frequency (\(\omega\)) and angular wave number (\(k\)). The formula to calculate wave speed \(v\) is:
  • \(v = \frac{\omega}{k}\)
This simple, yet powerful formula gives us valuable insight into the wave's behavior. In our exercise, the angular frequency is 30 s\(^{-1}\) and the wave number is 2 m\(^{-1}\). By plugging these values into our formula, we find:
  • \(v = \frac{30}{2} = 15 \, \text{m/s}\)
The wave speed on the string is therefore 15 meters per second. This means any point on the wave moves 15 meters in a second. By understanding wave speed, students can better grasp the dynamics of waves on strings.
Tension in a String
Tension in a string is a force that stretches or compresses the string. It's crucial because it affects how waves travel along it. Picture holding a tightly stretched rope with a friend and wiggling it; the tension greatly influences the wave propagation. For a string with waves, the relationship between wave speed \(v\), tension \(T\), and linear density \(\mu\) is:
  • \(v = \sqrt{\frac{T}{\mu}}\)
In this equation, higher tension results in a higher wave speed. To find the tension in the string based on known values, we rearrange the formula:
  • \(T = \mu v^2\)
By applying our exercise's values, where linear density \(\mu\) is \(1.6 \times 10^{-4} \, \text{kg/m}\) and wave speed \(v\) is 15 m/s:
  • \(T = 1.6 \times 10^{-4} \, \text{kg/m} \times (15 \, \text{m/s})^2 = 3.6 \, \text{N}\)
Thus, the string must be under 3.6 Newtons of tension. Understanding how tension interacts with wave properties helps students visualize wave dynamics more intuitively.
Angular Frequency
Angular frequency denotes how quickly a wave oscillates or cycles through its motion. The traditional frequency, in Hertz (cycles per second), describes the same concept. However, angular frequency, measured in radians per second, is more convenient for mathematical calculations involving waves (especially if you're dealing with wave equations). The angular frequency \(\omega\) is connected to the traditional frequency \(f\) by the equation:
  • \( \omega = 2\pi f \)
It signifies how many radians of angular "rotation" the wave progresses in a second. In our example, the wave has \( \omega = 30 \, \text{s}^{-1} \). By understanding the angular frequency, we get a clearer picture of the wave's oscillation speed, which helps in many physical contexts, such as sound waves and light waves. Remember, more "spin" or oscillation (higher \(\omega\)) typically means the wave is carrying more energy or information.
Linear Density
Linear density is a measure of how much mass is distributed along a string per unit length, represented by \(\mu\). Think of it as the "concentration" of mass in a line. Just like a heavier chain swings differently than a lighter one, a higher linear density changes how waves travel. It's measured in units like kilograms per meter (kg/m).
In wave mechanics, linear density plays a vital role when calculating wave speed and tension, impacting how quickly or slowly waves propagate along the string. With our exercise example:
  • Linear density \(\mu = 1.6 \times 10^{-4} \, \text{kg/m}\)
This value means that every meter of string weighs 0.00016 kilograms. The linear density parameter helps us compute not only the wave speed through the equation \(v = \sqrt{\frac{T}{\mu}}\) but also influences wave dynamics and energy transmission. Understanding the linear density is key to predicting and adjusting the wave behavior on strings.

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Most popular questions from this chapter

A generator at one end of a very long string creates a wave given by $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x+\left(8.00 \mathrm{~s}^{-1}\right) t\right] $$ and a generator at the other end creates the wave $$ y=(6.0 \mathrm{~cm}) \cos \frac{\pi}{2}\left[\left(2.00 \mathrm{~m}^{-1}\right) x-\left(8.00 \mathrm{~s}^{-1}\right) t\right] $$ Calculate the (a) frequency, (b) wavelength, and (c) speed of each wave. For \(x \geq 0\), what is the location of the node having the (d) smallest, (e) second smallest, and (f) third smallest value of \(x\) ? For \(x \geq 0\), what is the location of the antinode having the (g) smallest, (h) second smallest, and (i) third smallest value of \(x\) ?

A human wave. During sporting events within large, densely packed stadiums, spectators will send a wave (or pulse) around the stadium (Fig. 16-28). As the wave reaches a group of spectators, they stand with a cheer and then sit. At any instant, the width \(w\) of the wave is the distance from the leading edge (people are just about to stand) to the trailing edge (people have just sat down). Suppose a human wave travels a distance of 853 seats around a stadium in \(39 \mathrm{~s}\), with spectators requiring about \(1.8 \mathrm{~s}\) to respond to the wave's passage by standing and then sitting. What are (a) the wave speed \(v\) (in seats per second) and (b) width \(w\) (in number of seats)?

A \(120 \mathrm{~cm}\) length of string is stretched between fixed supports. What are the (a) longest, (b) second longest, and (c) third longest wavelength for waves traveling on the string if standing waves are to be set up? (d) Sketch those standing waves.

The equation of a transverse wave on a string is $$ y=(2.0 \mathrm{~mm}) \sin \left[\left(20 \mathrm{~m}^{-1}\right) x-\left(600 \mathrm{~s}^{-1}\right) t\right] $$ The tension in the string is \(15 \mathrm{~N}\). (a) What is the wave speed? (b) Find the linear density of this string in grams per meter.

The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: $$ \begin{array}{l} y_{1}(x, t)=(6.00 \mathrm{~mm}) \sin (4.00 \pi x-400 \pi t) \\ y_{2}(x, t)=(6.00 \mathrm{~mm}) \sin (4.00 \pi x+400 \pi t) \end{array} $$ with \(x\) in meters and \(t\) in seconds. An antinode is located at point \(A\). In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?

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