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Prove that \(x(t)=\cos \omega t\) oscillates with a frequency \(v=\omega / 2 \pi .\) Prove that \(x(t)=A \cos \omega t+B \sin \omega t\) oscillates with the same frequency, \(\omega / 2 \pi\)

Short Answer

Expert verified
Both functions oscillate with frequency \(v = \omega / 2\pi\).

Step by step solution

01

Understand the Oscillation Formula for Cosine

The function \(x(t) = \cos \omega t\) is a simple harmonic function. The general form of a harmonic function is \(x(t) = \cos(2\pi v t)\), where \(v\) is the frequency. By comparing it with \(x(t) = \cos \omega t\), we identify that \(\omega\) is equivalent to \(2\pi v\). Thus, the frequency \(v\) is given by \(v = \omega / 2\pi\).
02

Consider the Combination of Cosine and Sine

The function \(x(t) = A \cos \omega t + B \sin \omega t\) can be expressed as a linear combination of cosine and sine functions. These two functions individually are also harmonic functions with the frequency \(v = \omega / 2\pi\).
03

Convert to a Single Harmonic Function

Any linear combination of \(A \cos \omega t + B \sin \omega t\) can be rewritten as \(C \cos(\omega t + \phi)\), where \(C\) and \(\phi\) are constants. The single harmonic function \(C \cos(\omega t + \phi)\) clearly shows the oscillatory nature with frequency \(v = \omega / 2\pi\).
04

Apply Trigonometric Identities

To confirm the conversion, use trigonometric identities where \(C = \sqrt{A^2 + B^2}\) and \(\phi = \tan^{-1}(B/A)\). So, \(A \cos \omega t + B \sin \omega t = \sqrt{A^2 + B^2} \cos(\omega t + \phi)\), proving it is indeed a harmonic function with the same frequency, \(v = \omega / 2\pi\).

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Key Concepts

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

Frequency Calculation
In the context of harmonic oscillations, frequency represents how often a motion repeats itself. Frequency is linked to the oscillation period, which is the time taken for one complete cycle of motion. When dealing with the expression \(x(t) = \cos \omega t\), the term \(\omega\) is known as the angular frequency. Angular frequency is different from regular frequency by a factor of \(2\pi\). The relationship is given by the formula:
  • \(v = \frac{\omega}{2\pi}\)
This equation means that the regular frequency \(v\) is derived by dividing the angular frequency \(\omega\) by \(2\pi\). This conversion is crucial because regular frequency is typically measured in Hertz (cycles per second), and it tells us how many complete cycles occur each second. Knowing how to calculate the frequency helps to understand the oscillatory behavior of the function over time.
In a function like \(x(t) = A \cos \omega t + B \sin \omega t\), both components oscillate with the same frequency \(v = \omega/2\pi\) as they are based on the same angular frequency \(\omega\). Understanding frequency is key to analyzing harmonic oscillations.
Trigonometric Identities
Trigonometric identities play a vital role in simplifying and transforming harmonic functions. These identities are tools that allow us to rewrite expressions in different forms. When dealing with expressions like \(x(t) = A \cos \omega t + B \sin \omega t\), we use trigonometric identities for transformation.
A crucial identity used here is the one that combines sine and cosine terms into a single cosine expression:
  • \(A \cos \omega t + B \sin \omega t = C \cos(\omega t + \phi)\)
Where \(C = \sqrt{A^2 + B^2}\) and \(\phi = \tan^{-1}(B/A)\). These transformations simplify the analysis of oscillatory functions by expressing them in a single, clear form. Using these identities not only helps in simplifying expressions but also makes it easier to identify the amplitude and phase of the oscillation. Rather than dealing with two separate terms, we now have a single cosine term, which reflects the same oscillatory behavior and frequency.
Harmonic Functions
Harmonic functions are mathematical expressions that describe oscillatory motion. These functions are important because they model many physical phenomena, such as sound waves and light waves. A basic example is \(x(t) = \cos \omega t\), which represents simple harmonic motion.
This type of function shows a repeating pattern over time, making it periodic. The function repeats its values in regular intervals, known as periods. Another form of harmonic function is given by combining sine and cosine terms: \(x(t) = A \cos \omega t + B \sin \omega t\). Such a combination can be turned into another harmonic function \(C \cos(\omega t + \phi)\), capturing the same periodic behavior.
Harmonic functions are defined by their oscillatory pattern, amplitude, and frequency. The amplitude tells us how large the oscillations are, while the frequency tells how often they occur. When analyzing physical systems, harmonic functions provide an efficient way to represent and predict periodic behaviors.

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Most popular questions from this chapter

Solve the following differential equations: a. \(\frac{d^{2} y}{d x^{2}}-4 y=0 \quad y(0)=2 ; \frac{d y}{d x}(\text { at } x=0)=4\) b. \(\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}+6 y=0 \quad y(0)=-1 ; \frac{d y}{d x}(\text { at } x=0)=0\) c. \(\frac{d y}{d x}-2 y=0 \quad y(0)=2\)

Prove that $$y(x, t)=A \sin \left[\frac{2 \pi}{\lambda}(x-v t)\right]$$ is a wave of wavelength \(\lambda\) and frequency \(v=v / \lambda\) traveling to the right with a velocity \(v\).

Find the general solutions to the following differential equations. a. \(\frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+3 y=0\) b. \(\frac{d^{2} y}{d x^{2}}+6 \frac{d y}{d x}=0\) c. \(\frac{d y}{d x}+3 y=0\) d. \(\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}-y=0\) e. \(\frac{d^{2} y}{d x^{2}}-3 \frac{d y}{d x}+2 y=0\)

Prove that the number of nodes for a vibrating string clamped at both ends is \(n-1\) for the \(n\) th harmonic.

This problem is the extension of Problem 2-9 to two dimensions. In this case, the particle is constrained to move freely over the surface of a rectangle of sides \(a\) and \(b\). The Schrödinger equation for this problem is $$ \frac{\partial^{2} \psi}{\partial x^{2}}+\frac{\partial^{2} \psi}{\partial y^{2}}+\left(\frac{8 \pi^{2} m E}{h^{2}}\right) \psi(x, y)=0 $$ with the boundary conditions $$ \begin{aligned} &\psi(0, y)=\psi(a, y)=0 \quad \text { for all } y, \quad 0 \leq y \leq b \\ &\psi(x, 0)=\psi(x, b)=0 \quad \text { for all } x, \quad 0 \leq x \leq a \end{aligned} $$ Solve this equation for \(\psi(x, y)\), apply the boundary conditions, and show that the energy is quantized according to $$ E_{n_{x} n_{y}}=\frac{n_{x}^{2} h^{2}}{8 m a^{2}}+\frac{n_{y}^{2} h^{2}}{8 m b^{2}} \quad \begin{array}{ll} n_{x}=1,2,3, \ldots \\ n_{y}=1,2,3, \ldots \end{array} $$

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