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Given the wave functions \(y_{1}(x, t)=A \sin (k x-\omega t)\) \(y_{2}(x, t)=A \sin (k x-\omega t+\phi) \quad\) with \(\quad \phi \neq \frac{\pi}{2},\) show that \(y_{1}(x, t)+y_{2}(x, t)\) is a solution to the linear wave equation with a wave velocity of \(v=\sqrt{\frac{\omega}{k}}\).

Short Answer

Expert verified
The sum of the wave functions \(Y(x,t) = A \sin(kx - \omega t) + A \sin(kx - \omega t + \phi)\) can be simplified using the sum-to-product trigonometric identity, resulting in \(Y(x,t) = 2A\sin(kx - \omega t + \frac{\phi}{2})\cos(\frac{\phi}{2})\). By calculating the second derivatives with respect to time and position, we find that \(Y(x,t)\) satisfies the linear wave equation with wave velocity \(v = \sqrt{\frac{\omega}{k}}\), thus proving the solution.

Step by step solution

01

Combine the wave functions y1 and y2.

Sum the given wave functions y1(x,t) and y2(x,t) to get a new wave function Y(x,t) representing the superposition of the two waves: \( Y(x, t) = y_{1}(x, t) + y_{2}(x, t) = A \sin(kx - \omega t) + A \sin(kx - \omega t + \phi) \). #Step 2: Simplifying using trigonometric Identity
02

Apply the trigonometric identity to simplify Y(x,t).

Apply the sum-to-product trigonometric identity to simplify Y(x,t), which states that: \( \sin(A) + \sin(B) = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2}) \). In our case, \(A = kx - \omega t\) and \(B = kx - \omega t + \phi\), thus: \( Y(x, t) = 2A \sin(\frac{(kx - \omega t) + (kx - \omega t + \phi)}{2})\cos(\frac{(kx - \omega t) - (kx - \omega t + \phi)}{2}) \). Simplifying, we get: \( Y(x, t) = 2A\sin(kx - \omega t + \frac{\phi}{2})\cos(\frac{\phi}{2}) \). #Step 3: Linear wave equation#
03

Write down the linear wave equation.

The linear wave equation for a wave in one dimension is given by: \( \frac{\partial^2Y}{\partial t^2} = v^2 \frac{\partial^2Y}{\partial x^2} \). To show that Y(x,t) is a solution, we need to compute its second derivatives with respect to time t and position x, and then check if both sides of the equation are equal. #Step 4: Second time derivative of Y#
04

Calculate the second time derivative of Y(x,t).

To find the second time derivative, take the first derivative with respect to t, then take the derivative of the resulting expression again with respect to t: \( \frac{\partial Y}{\partial t} = -2A\omega\sin(\frac{\phi}{2})\cos(kx - \omega t + \frac{\phi}{2}) \). \( \frac{\partial^2 Y}{\partial t^2} = 2A\omega^2\sin(\frac{\phi}{2})\sin(kx - \omega t + \frac{\phi}{2}) \). #Step 5: Second position derivative of Y#
05

Calculate the second position derivative of Y(x,t).

Similar to Step 4, take the first derivative with respect to x, and take the derivative of the resulting expression again with respect to x: \( \frac{\partial Y}{\partial x} = 2Ak\sin(\frac{\phi}{2})\cos(kx - \omega t + \frac{\phi}{2}) \). \( \frac{\partial^2 Y}{\partial x^2} = -2Ak^2\sin(\frac{\phi}{2})\sin(kx - \omega t + \frac{\phi}{2}) \). #Step 6: Check if Y is a solution to the linear wave equation #
06

Verify if Y(x,t) satisfies the linear wave equation.

Now, we will plug our results from Steps 4 and 5 into the linear wave equation and check if both sides are equal: \( 2A\omega^2\sin(\frac{\phi}{2})\sin(kx - \omega t + \frac{\phi}{2}) = v^2 (-2Ak^2\sin(\frac{\phi}{2})\sin(kx - \omega t + \frac{\phi}{2})) \). Cancel \(2A\sin(\frac{\phi}{2})\sin(kx - \omega t + \frac{\phi}{2})\) from both sides and solve for v: \( \frac{\omega^2}{k^2} = v^2 \). Finally, taking the square root of both sides, we get: \( v = \sqrt{\frac{\omega}{k}} \). Since we have verified that Y(x,t) satisfies the linear wave equation with wave velocity given by \(v = \sqrt{\frac{\omega}{k}}\), the solution is proven.

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Key Concepts

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

Understanding the Linear Wave Equation
The linear wave equation is a fundamental concept in physics which describes how wave functions, such as those for sound or light, propagate through various media. The linear wave equation for a one-dimensional wave can be expressed as
\[\begin{equation}\frac{\partial^2 Y}{\partial t^2} = v^2 \frac{\partial^2 Y}{\partial x^2}. \end{equation}\]
Where \(Y(x, t)\) represents the displacement of the wave, \(v\) is the wave velocity, \(x\) is the position, and \(t\) is the time. This equation says that the acceleration of the wave function, as denoted by the second time derivative, is proportional to the curvature of the wave function, as shown by the second spatial derivative. Essential to solving the linear wave equation is the concept of partial differentiation, which involves taking the derivative of a multivariable function with respect to one variable while keeping others constant.
Through step-by-step partial differentiation and substitution, one can demonstrate that a superposed wave function indeed satisfies the linear wave equation provided it results in a consistent velocity term. This approach not only confirms the validity of the superposed wave under certain conditions but also reinforces the notion that linear superposition applies to waves governed by the linear wave equation.
Trigonometric Identities and Wave Superposition
Trigonometric identities are valuable tools in simplifying the expressions involving trigonometric functions, which are common in wave equations. These identities transform complex trigonometric expressions into more manageable forms. For instance, the sum-to-product identities which transform the sum of two sine functions into a product form are particularly useful in analyzing the superposition of waves.
In the context of waves, superposition is when two or more waves overlap and combine into a new wave pattern. Mathematically, this combination can lead to complex expressions which can be gracefully simplified using trigonometric identities. As demonstrated in the exercise solution, the identity \[\begin{equation}\sin(A) + \sin(B) = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})\end{equation}\]
allows the sum of the wave functions \(y_1(x, t)\) and \(y_2(x, t)\) to be expressed in a form that makes partial differentiation much simpler. This simplification is crucial in proving whether the superposed wave satisfies the linear wave equation, ultimately enabling further analysis of wave behavior.
Partial Differentiation in Physics
Partial differentiation is a technique used in mathematics and physics to deal with functions of multiple variables. It is essential when describing phenomena where a change in one variable may affect the outcome independently of other variables. In physics, particularly wave mechanics, partial derivatives help in understanding how waves change over time and space.
When dealing with the linear wave equation, partial differentiation allows us to find the rate of change of the wave function with respect to time and space. Step by step, as outlined in the exercise solution, one can take the first and second partial derivatives with respect to time (\(t\)) and position (\(x\)) to analyze the wave motion. The calculations provided in the step-by-step solution showcase the use of partial differentiation to verify the wave equation. It’s a cornerstone for proving that a given function is indeed a proper solution to the wave equation, as both sides of the equation must balance when the second derivatives are plugged in. Understanding both the process and the application of partial differentiation in wave mechanics is key for students to grasp not only wave superposition but also the broader implications of wave behavior in various physical systems.

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Most popular questions from this chapter

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