Chapter 16: Problem 14
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.
Short Answer
Step by step solution
Identify the Wave Equation
Calculate the Wave Speed
Find the Linear Density
Summary of Results
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Transverse Wave
In the given equation, this concept is key to understanding how the wave propagates along the string. The equation provided demonstrates the sine wave form that is characteristic of transverse waves, where the amplitude of the wave ( 2.0 ext{ mm} ) describes the maximum displacement from the rest position.
When dealing with transverse waves on strings, it is crucial to consider factors like the medium (string in this case) and its properties (like tension and linear density) as they directly influence wave characteristics.
Linear Density
In this problem, linear density affects how the string responds to tension. A higher linear density means more mass, which can affect how fast vibrations travel through the string.
From the calculation, the linear density μ was found using the formula v = ext{ sqrt(T/µ) } . Converting from ext{ kg/m } to ext{ g/m } provides values that are easier for practical experiments or applications.
Wave Equation
In this case, y = 2.0 ext{ mm} ext{ sin }[(20 ext{ m}^{-1})x - (600 ext{ s}^{-1})t] describes the displacement y of a point on the string over time t and position x .
Understanding the wave equation is essential as it encapsulates all necessary parameters that tell us about wave motion in the context of mechanics.
Wave Number
For this wave equation, k = 20 ext{ m}^{-1} implies that every meter of the wave propagation consists of 20 complete wave cycles.
By knowing the wave number, you can determine important wave properties like wavelength and how compactly the wave oscillates in space, which are pivotal in understanding how waves behave in different media.
Angular Frequency
In our equation, ω = 600 ext{ s}^{-1} , indicating that the wave oscillates 600 full cycles per second. This fast oscillation contributes to the dynamics and energy of the wave.
Angular frequency is a crucial parameter because it helps in calculating the wave speed using the formula v = ω/k , where k is the wave number. Understanding angular frequency allows for deeper insights into the temporal dynamics of the wave.