Math Problem Statement

Which of the following is a solution of the one-dimensional wave equation ∂²u/∂x² = (1/c²)∂²u/∂t², with c = 5, for the given functions u(x, t)?

Solution

The given problem is about finding a solution to the one-dimensional wave equation:

2ux2=1c22ut2\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}

where c=5c = 5. The options provide different functions u(x,t)u(x, t), and we need to determine which of these functions satisfy the wave equation for the given value of cc.

Step-by-Step Analysis

To determine if a function is a solution to the wave equation, it must satisfy the equation:

2ux2=1c22ut2\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}

Let's analyze each of the given options:

  1. Option (I): u(x,t)=sin(x+2t)+sin(x2t)u(x, t) = \sin(x + 2t) + \sin(x - 2t)

    • Here, the wave speed would be determined by the coefficients of xx and tt inside the sine functions.
    • For a solution to satisfy the wave equation with c=5c = 5, the coefficient ratio should be such that coefficient of xcoefficient of t=c=5\frac{\text{coefficient of } x}{\text{coefficient of } t} = c = 5.
    • For both terms sin(x+2t)\sin(x + 2t) and sin(x2t)\sin(x - 2t), the coefficient of tt is 2, and the coefficient of xx is 1. Thus, this does not satisfy c=5c = 5, so (I) is not a solution.
  2. Option (II): u(x,t)=sin2(5xt)u(x, t) = \sin^2(5x - t)

    • The argument 5xt5x - t suggests a wave speed c=51=5c = \frac{5}{1} = 5.
    • This matches the given wave speed c=5c = 5. Therefore, (II) could be a solution.
    • However, note that this is sin2(5xt)\sin^2(5x - t), not a simple sine function. The second derivatives of sin2\sin^2 still satisfy the wave equation due to the chain rule, as it represents a combination of two sinusoidal functions with the correct wave speed.
  3. Option (III): u(x,t)=(9x+5t)6u(x, t) = (9x + 5t)^6

    • The function is a polynomial, not a sinusoidal or cosine function.
    • For this to satisfy the wave equation, it should represent a traveling wave form, which it does not. Therefore, (III) is not a solution.
  4. Option (IV): u(x,t)=cos2(3x15t)u(x, t) = \cos^2(3x - 15t)

    • The argument 3x15t3x - 15t suggests a wave speed c=153=5c = \frac{15}{3} = 5.
    • This matches the given wave speed c=5c = 5. Therefore, (IV) could be a solution.
    • The cos2\cos^2 term also satisfies the wave equation similar to the sine squared term due to the chain rule.

Conclusion

Based on the analysis:

  • (II) and (IV) both satisfy the wave equation for c=5c = 5.
  • Therefore, the correct answer is:

(II) and (IV) only.

Would you like to know more about how the wave equation works or how to verify these solutions in detail?

Related Questions:

  1. What is the general solution to the one-dimensional wave equation?
  2. How do you determine if a function satisfies a partial differential equation?
  3. Why do both sine squared and cosine squared functions satisfy the wave equation?
  4. What are some physical interpretations of the wave equation?
  5. How would you solve the wave equation using separation of variables?

Tip: Understanding the form of solutions to the wave equation is key in physics and engineering, as it represents many physical phenomena like sound and light waves.

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Math Problem Analysis

Mathematical Concepts

Partial Differential Equations
Wave Equation
Solution Verification

Formulas

∂²u/∂x² = (1/c²)∂²u/∂t²
Wave speed formula c = distance/time

Theorems

Wave Equation
Solution of PDEs in terms of traveling waves

Suitable Grade Level

Undergraduate (Advanced High School / Early University)