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Verify by direct substitution that the function \(\phi(t)=A \sin (\omega t)+B \cos (\omega t)\) of (1.56) is a solution of the second-order differential equation \((1.55), \ddot{\phi}=-\omega^{2} \phi .\) (since this solution involves two arbitrary constants - the coefficients of the sine and cosine functions \(-\) it is in fact the general solution.)

Short Answer

Expert verified
The function \(\phi(t)\) satisfies the differential equation by substitution.

Step by step solution

01

Differentiate the given function \( \phi(t) \)

First, find the first derivative of \( \phi(t) = A \sin(\omega t) + B \cos(\omega t) \) with respect to \( t \). Using the standard derivative rules, the derivative is \( \phi'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t) \).
02

Differentiate again to find \( \ddot{\phi} \)

Differentiate \( \phi'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t) \) to find the second derivative. Utilizing the standard derivatives again, the result is \( \ddot{\phi}(t) = -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) \).
03

Substitute into the differential equation

Substitute \( \ddot{\phi}(t) = -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) \) into the equation \( \ddot{\phi} = -\omega^2 \phi \). Check if both sides are equal: \(-A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) = -\omega^2 [A \sin(\omega t) + B \cos(\omega t)]\), which simplifies to the same expression as the left side.

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Key Concepts

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

General Solution
A general solution to a differential equation is a formula that encompasses all possible solutions of that equation. This means it contains arbitrary constants that can be adjusted to fit specific boundary conditions or initial values. In our problem, the general solution involves the equation \[ \phi(t) = A \sin(\omega t) + B \cos(\omega t) \] Here, \( A \) and \( B \) are the arbitrary constants. These are adjustable and allow the solution to cater to any initial conditions presented by a problem. The function \( \phi(t) \) being a solution means every combination of \( A \) and \( B \) satisfies the equation \( \ddot{\phi} = -\omega^2 \phi \). Therefore, it provides a comprehensive set of solutions for different scenarios. Having such a general form is useful because it provides flexibility in solving differential equations related to waves, oscillations, and other periodic phenomena.
Sine and Cosine Functions
Sine and cosine functions are fundamental in the study of waves and oscillations. In the equation \( \phi(t) = A \sin(\omega t) + B \cos(\omega t) \), both functions contribute to the behavior of the solution. These trigonometric functions are periodic and continuous, characteristics that make them ideal for describing oscillatory systems like springs and pendulums.

Key characteristics include:
  • Periodicity: Both sine and cosine repeat their values in cyclic patterns, having a period of \( 2\pi \).
  • Amplitude Modulation: The constants \( A \) and \( B \) determine the amplitude of each function, reflecting the maximum extent of oscillation.
  • Phase Relationship: Sine and cosine functions are phase-shifted by \( \pi/2 \) radians; this is crucial in modeling waves where such phase shifts create interference patterns.
These functions have wide applications, not only in simple harmonic motion but also in Fourier analysis, which breaks down more complex waves into sums of sine and cosine components.
Differentiation
Differentiation is a process of finding the rate at which a function is changing. In the context of differential equations, differentiation helps us understand how quantities evolve over time. For the function \( \phi(t) = A \sin(\omega t) + B \cos(\omega t) \), differentiation involves:
  • First Derivative: Differentiating \( \phi(t) \) with respect to \( t \) gives \( \phi'(t) = A \omega \cos(\omega t) - B \omega \sin(\omega t) \). This represents the instantaneous rate of change of the function.
  • Second Derivative: Further differentiation gives \( \ddot{\phi}(t) = -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) \). The second derivative is crucial as it reveals how the rate of change itself changes, often related to acceleration and force in physics.
Differentiation is indispensable in solving differential equations since it allows us to work backwards from a general solution and verify if it satisfies the given conditions.
Direct Substitution
Direct substitution is a technique used to verify if a particular solution satisfies a given equation. It involves plugging the solution and its derivatives back into the equation to ensure consistency and correctness. In our case:
  • Start by differentiating \( \phi(t) \) to find both \( \phi'(t) \) and \( \ddot{\phi}(t) \).
  • Substitute \( \ddot{\phi}(t) = -A \omega^2 \sin(\omega t) - B \omega^2 \cos(\omega t) \) into the differential equation \( \ddot{\phi} = -\omega^2 \phi \).
  • Ensure that this substitution leads to equivalent expressions on both sides of the equation.
Successful substitution implies that \( \phi(t) \) is indeed a valid solution to the differential equation. This straightforward process confirms the relationship between the solution and the equation, providing confidence in the derived general solution.

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Most popular questions from this chapter

In case you haven't studied any differential equations before, I shall be introducing the necessary ideas as needed. Here is a simple excercise to get you started: Find the general solution of the firstorder equation \(d f / d t=f\) for an unknown function \(f(t) .\) [There are several ways to do this. One is to rewrite the equation as \(d f / f=d t\) and then integrate both sides.] How many arbitrary constants does the general solution contain? [Your answer should illustrate the important general theorem that the solution to any \(n\) th-order differential equation (in a very large class of "reasonable" equations) contains \(n\) arbitrary constants.]

The position of a moving particle is given as a function of time \(t\) to be $$\mathbf{r}(t)=\hat{\mathbf{x}} b \cos (\omega t)+\hat{\mathbf{y}} c \sin (\omega t)$$ where \(b, c,\) and \(\omega\) are constants. Describe the particle's orbit.

Prove that if \(\mathbf{v}(t)\) is any vector that depends on time (for example the velocity of a moving particle) but which has constant magnitude, then \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t) .\) Prove the converse that if \(\dot{\mathbf{v}}(t)\) is orthogonal to \(\mathbf{v}(t),\) then \(|\mathbf{v}(t)|\) is constant. [Hint: Consider the derivative of \(\mathbf{v}^{2}\).] This is a very handy result. It explains why, in two- dimensional polars, \(d \hat{\mathbf{r}} / d t\) has to be in the direction of \(\hat{\boldsymbol{\phi}}\) and vice versa. It also shows that the speed of a charged particle in a magnetic field is constant, since the acceleration is perpendicular to the velocity.

A parallelepiped (a six-faced solid with opposite faces parallel) has one corner at the origin \(O\) and the three edges that emanate from \(O\) defined by vectors \(\mathbf{a}, \mathbf{b}, \mathbf{c}\) Show that the volume of the parallelepiped is \(|\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c})|\)

[Computer] The differential equation (1.51) for the skateboard of Example 1.2 cannot be solved in terms of elementary functions, but is easily solved numerically. (a) If you have access to software, such as Mathematica, Maple, or Matlab, that can solve differential equations numerically, solve the differential equation for the case that the board is released from \(\phi_{\mathrm{o}}=20\) degrees, using the values \(R=5 \mathrm{m}\) and \(g=9.8 \mathrm{m} / \mathrm{s}^{2} .\) Make a plot of \(\phi\) against time for two or three periods. ( \(\mathbf{b}\) ) On the same picture, plot the approximate solution (1.57) with the same \(\phi_{\mathrm{o}}=20^{\circ}\) Comment on your two graphs. Note: If you haven't used the numerical solver before, you will need to learn the necessary syntax. For example, in Mathematica you will need to learn the syntax for "NDSolve”and how to plot the solution that it provides. This takes a bit of time, but is something that is very well worth learning.

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